"Babeş-Bolyai" University

Faculty of Mathematics and Physics

Research Seminars

Seminar on Functional Analysis and Numerical Methods

Preprint Nr.1, 1987, pp.121-129

Error estimations in the numerical solving of systems of equations in metric spaces

Ion Păvăloiu

Let us designate by $\left(X_{i},\rho_{i}\right),\ i=1,2$ two complete metric spaces and by $X=X_{1}\times X_{2}$ the Cartesian product of these spaces.

We designate by $F_{1}:X\rightarrow X_{1}$ And $F_{2}:X\rightarrow X_{2}$ two applications and we consider the following system of equations:

(1)		$\displaystyle x_{1}$	$\displaystyle=F_{1}\left(x_{1},x_{2}\right)$
	$\displaystyle x_{2}$	$\displaystyle=F_{2}\left(x_{1},x_{2}\right),\qquad\left(x_{1},x_{2}\right)\in X.$

For the resolution of the system of equations ( 1 ) we consider the following iterative process, of the Gauss-Seidel type:

(2)		$\displaystyle x_{1}^{\left(n+1\right)}$	$\displaystyle=F_{1}\left(x_{1}^{\left(n\right)},x_{2}^{\left(n\right)}\right)$
	$\displaystyle x_{2}^{\left(n+1\right)}$	$\displaystyle=F_{2}\left(x_{1}^{\left(n+1\right)},x_{2}^{\left(n\right)}\right% ),\qquad n=0,1,\ldots,\left(x_{1}^{\left(0\right)},x_{2}^{\left(0\right)}% \right)\in X.$

Concerning the convergence of sequences we have the following Theorem [ 1 ] :

THEOREM 1 .

If the applications $F_{1}\$ And $F_{2}$ check the conditions

	$\displaystyle\rho_{1}\left(F_{1}\left(x_{1},y_{1}\right),F_{1}\left(x_{2},y_{2% }\right)\right)$	$\displaystyle\leq\alpha\rho_{1}\left(x_{1},x_{2}\right)+\beta\rho_{2}\left(y_{% 1},y_{2}\right),$
	$\displaystyle\rho_{2}\left(F_{2}\left(x_{1},y_{1}\right),F_{2}\left(x_{2},y_{2% }\right)\right)$	$\displaystyle\leq a\rho_{1}\left(x_{1},x_{2}\right)+b\rho_{2}\left(y_{1},y_{2}\right)$

for all $\left(x_{1},y_{1}\right),\left(x_{2},y_{2}\right)\in X$ , Or $\alpha,\beta,a$ And $b$ are nonnegative real numbers;

If the numbers $\alpha,\beta,a$ And $b$ check the relationships

	$\displaystyle\alpha+b+a\beta$	$\displaystyle<2$
	$\displaystyle\left(1-\alpha\right)\left(1-b\right)$	$\displaystyle>a\beta,\qquad b>0,\ \ \alpha>0$

then the system ( 1 ) admits a single solution $\left(\bar{x}_{1},\bar{x}_{2}\right)\in X\$ and the sequels $\ \big{(}x_{1}^{\left(n\right)}\big{)}_{n=0}^{\infty},\ \big{(}x_{2}^{\left(n% \right)}\big{)}_{n=0}^{\infty}$ are convergent:

\lim_{n\rightarrow\infty}x_{1}^{\left(n\right)}=\bar{x}_{1},\quad\lim_{n% \rightarrow\infty}x_{2}^{\left(n\right)}=\bar{x}_{2}.

Demonstration.

We designate by $\left(f_{n}\right)_{n=0}^{\infty}$ And $\left(g_{n}\right)_{n=0}^{\infty}$ two sequences of non-negative numbers whose elements verify the relations

(3)		$\displaystyle f_{n}$	$\displaystyle\leq\alpha f_{n-1}+\beta g_{n-1}$
	$\displaystyle g_{n}$	$\displaystyle\leq af_{n}+bg_{n-1},\qquad n=1,2,\ldots$

We associate with relations ( 3 ) the system of equations in the unknowns $h$ And $k$ following

(4)		$\displaystyle\alpha+\beta h$	$\displaystyle=hk$
	$\displaystyle ak+b$	$\displaystyle=nk$

We will subsequently show that the system ( 4 ) admits a real solution $\left(h_{i},k_{i}\right)\$ for which $0\leq h_{i}k_{i}<1$ if and only if the numbers $a,b,\alpha$ And $\beta$ fulfill the conditions of Theorem 1 .

We assume that the system ( 4 ) admits the solutions $\left(h_{i},k_{i}\right)$ , $i=1,2$ for which the conditions $0<h_{i}k_{i}<1$ are fulfilled. We immediately check that the following equations result from the system ( 4 ):

(5)		$\displaystyle\beta h^{2}-\left(b+\beta a-\alpha\right)h-\alpha a$	$\displaystyle=0$
	$\displaystyle ak^{2}-\left(\alpha+\beta a-b\right)k-\beta b$	$\displaystyle=0$

And

(6)

p^{2}-\left(b+\beta a+\alpha\right)p-b\alpha=0

where we have designated by $p$ the product $h k .$ Equations ( 5 ) show us that the solutions $\left(h_{i},k_{i}\right),\ i=1,2$ of the system ( 4 ) are real and of the equation ( 6 ) and the conditions $0<p_{i}<1,$ $i=1,2$ it results $f(0)>0$ , $f(1)>0$ And $b+\beta a+\alpha<2,$ Or $f\left(p\right)=p^{2}-\left(b+\beta a+\alpha\right)p+b\alpha$ . It is obvious that the condition $f(0)>0$ is filled because $\alpha>0,b>0$ , $f(1)>0$ And $b+a\beta+\alpha<2$ represent the relations of the hypothesis of Theorem 1 .

If we now assume that the relations of the hypothesis of Theorem 1 are verified, then it is obvious that $f\left(1\right)>0,$ $f\left(0\right)>0$ And $b+a\beta+\alpha<2,$ which shows us that equation ( 6 ) has both positive roots and roots less than unity. Taking into account equations ( 5 ), we easily see that system ( 4 ) admits a solution $\left(h_{1},k_{1}\right)$ for which $h_{1}>0,\ k_{1}>0.$

We now show that if the elements of the sequences $\left(f_{n}\right)_{n=0}^{\infty}$ And $\left(g_{n}\right)_{n=0}^{\infty}$ verify the relations ( 3 ) where the numbers $\alpha,\beta,a$ And $b$ verify the hypothesis of Theorem 1 , then there exists a constant $C>0$ , independent of $n$ such that for each $n=1,2,\ldots$ relationships take place

(7)		$\displaystyle f_{n}$	$\displaystyle\leq Ch_{1}^{n-1}k_{1}^{n-1}$
	$\displaystyle g_{n}$	$\displaystyle\leq Ch_{1}^{n}k_{1}^{n-1}$

and the series $\sum_{i=0}^{\infty}f_{i}$ And $\sum_{i=0}^{\infty}g_{i}$ are convergent.

Let us designate by $C$ a real number that satisfies the inequality

(8)

C\geq\max\left\{\alpha f_{0}+\beta g_{0},\tfrac{af_{1}+bg_{0}}{h_{1}}\right\}

Or $\left(h_{1},k_{1}\right)$ is the positive solution of the system ( 4 ).

From ( 8 ) and ( 3 ) we deduce for $n=1$

f_{1}\leq C\ \ \ \ \text{and\quad}g_{1}\leq Ch_{1}

so relations ( 7 ) are verified for $n=1.$

We assume that the following inequalities hold:

f_{k-1}\leq C\ h_{1}^{k-2}\cdot k_{1}^{k-2},\qquad g_{k-1}\leq C\ h_{1}^{k-1}% \cdot k_{1}^{k-2},\qquad k=2,\ldots,n

From ( 3 ) and ( 4 ) we deduce

	$\displaystyle f_{n}$	$\displaystyle\leq\alpha f_{n-1}+\beta g_{n-1}\leq Ch_{1}^{n-2}k_{1}^{n-2}\left% (\alpha+\beta h_{1}\right)=Ch_{1}^{n-1}k_{1}^{n-1}$
	$\displaystyle g_{n}$	$\displaystyle\leq af_{n}+bg_{n-1}\leq C\ h_{1}^{n-1}k_{1}^{n-2}\left(ak_{1}+b% \right)=Ch_{1}^{n}k_{1}^{n-1}$

which shows us that relations ( 7 ) take place for all $n\in\mathbb{N}$ .

It is obvious that the series $\sum_{i=0}^{\infty}f_{i}$ And $\sum_{i=0}^{\infty}g_{i}$ are convergent, because from ( 7 ) it follows that they are bounded above by two geometric series with a ratio less than unity.

From ( 2 ) and the hypothesis of Theorem 1 we deduce the relations:

(9)		$\displaystyle\rho_{1}\left(x_{1}^{\left(n+1\right)},x_{1}^{\left(n\right)}\right)$	$\displaystyle\leq\alpha\rho_{1}\left(x_{1}^{\left(n\right)},x_{1}^{\left(n-1% \right)}\right)+\beta\rho_{2}\left(x_{2}^{\left(n\right)},x_{2}^{\left(n-1% \right)}\right)$
	$\displaystyle\rho_{2}\left(x_{2}^{\left(n+1\right)},x_{2}^{\left(n\right)}\right)$	$\displaystyle\leq a\rho_{1}\left(x_{1}^{\left(n+1\right)},x_{1}^{\left(n\right% )}\right)+b\rho_{2}\left(x_{2}^{\left(n\right)},x_{2}^{\left(n-1\right)}\right% ),\quad n=0,1,\ldots$

We now set in ( 9 ) $f_{i}=\rho_{1}\left(x_{1}^{\left(i+1\right)},x_{1}^{\left(i\right)}\right),g_{% i}=\rho_{2}\left(x_{2}^{\left(i+1\right)},x_{2}^{\left(i\right)}\right),$ $i=0,1,2,\ldots$ and we obtain the relations ( 3 ). Taking into account the hypotheses of Theorem 1 we deduce the relations

(10)		$\displaystyle\rho_{1}\left(x_{1}^{\left(i+1\right)},x_{1}^{\left(i\right)}\right)$	$\displaystyle\leq C\ h_{1}^{i-1}\cdot k_{1}^{i-1}$
	$\displaystyle\rho_{2}\left(x_{2}^{\left(i+1\right)},x_{2}^{\left(i\right)}\right)$	$\displaystyle\leq C\ h_{1}^{i}\cdot k_{1}^{i-1},\quad i=1,2,\ldots$

We will now show that suites $\left(x_{1}^{\left(n\right)}\right)_{n=0}^{\infty}$ And $\left(x_{2}^{\left(n\right)}\right)_{n=0}^{\infty}$ are convergent.

We have indeed

(11)		$\displaystyle\rho_{1}\left(x_{1}^{\left(n+s\right)},x_{1}^{\left(n\right)}\right)$	$\displaystyle\leq f_{n+s-1}+f_{n+s-2}+\ldots+f_{n}$
		$\displaystyle\leq p_{1}^{n-1}C\left(1+p_{1}+p_{1}^{2}+\ldots+p_{1}^{s-1}\right% )\leq\tfrac{Cp_{1}^{n-1}}{1-p_{1}}$

Or $p_{1}=h_{1}k_{1}$ .

We deduce in the same way

(12)

\rho_{2}\left(x_{2}^{\left(n+s\right)},x_{2}^{\left(n\right)}\right)\leq g_{n+% s-1}+g_{n+s-2}+\ldots+g_{n}\leq\tfrac{Ch_{1}p_{1}^{n-1}}{1-p_{1}}.

Taking into account that $0<p_{1}<1$ and the fact that spaces $X_{1}$ And $X_{2}$ are complete, it follows that the sequences $\big{(}x_{1}^{\left(n\right)}\big{)}_{n=0}^{\infty}$ And $\big{(}x_{2}^{\left(n\right)}\big{)}_{n=0}^{\infty}$ are convergent.

If we pose $\bar{x}_{1}=\lim\limits_{n\rightarrow\infty}x_{1}^{\left(n\right)}$ and $\bar{x}_{2}=\lim\limits_{n\rightarrow\infty}x_{2}^{\left(n\right)}$ then taking into account the continuity of applications $F_{1}$ And $F_{2}$ and passing to the limit in the equalities ( 2 ) when $n\rightarrow\infty$ it follows that $\left(\bar{x}_{1},\bar{x}_{2}\right)$ represents a solution of the system ( 1 ).

Regarding the uniqueness of the solution, we will assume by contradiction that the system ( 1 ) does not have a unique solution.

Let us designate by $\left(\bar{x}_{1},\bar{x}_{2}\right)$ And $\left(\bar{y}_{1},\bar{y}_{2}\right)$ two solutions of the system ( 1 ). It follows from the first condition of Theorem 1 :

	$\displaystyle\rho_{1}\left(\bar{x}_{1},\bar{y}_{1}\right)$	$\displaystyle\leq\alpha\rho_{1}\left(\bar{x}_{1},\bar{y}_{1}\right)+\beta\rho_% {2}\left(\bar{x}_{2},\bar{y}_{2}\right)$
	$\displaystyle\rho_{2}\left(\bar{x}_{2},\bar{y}_{2}\right)$	$\displaystyle\leq a\ \rho_{1}\left(\bar{x}_{1},\bar{y}_{1}\right)+b\rho_{2}% \left(\bar{x}_{2},\bar{y}_{2}\right)$

	$\displaystyle\left(1-\alpha\right)\rho_{1}\left(\bar{x}_{1},\bar{y}_{1}\right)$	$\displaystyle\leq\beta\rho_{2}\left(\bar{x}_{2},\bar{y}_{2}\right)$
	$\displaystyle\left(1-b\right)\rho_{2}\left(\bar{x}_{1},\bar{y}_{2}\right)$	$\displaystyle\leq a\rho_{1}\left(\bar{x}_{1},\bar{y}_{1}\right).$

We deduce from this

\rho_{1}\left(\bar{x}_{1},\bar{y}_{1}\right)\leq\tfrac{\beta\ a}{\left(1-% \alpha\right)\left(1-b\right)}\rho_{1}\left(\bar{x}_{1},\bar{y}_{1}\right)

And

\rho_{2}\left(\bar{x}_{2},\bar{y}_{2}\right)\leq\tfrac{\beta\ a}{\left(1-% \alpha\right)\left(1-b\right)}\rho_{2}\left(\bar{x}_{2},\bar{y}_{2}\right),

but $\frac{\beta\ a}{\left(1-\alpha\right)\left(1-b\right)}<1$ therefore the last inequalities take place if and only if $\bar{x}_{1}=\bar{y}_{1}$ And $\bar{x}_{2}=\bar{y}_{2}.$

It follows from ( 11 ) and ( 12 ) that $s\rightarrow\infty$

\rho_{1}\big{(}\bar{x}_{1},x_{1}^{\left(n\right)}\big{)}\leq\tfrac{C\cdot p_{1% }^{n-1}}{1-p_{1}}\qquad et\quad\rho_{2}\big{(}\bar{x}_{2},x_{2}^{\left(n\right% )}\big{)}\leq\tfrac{Ch_{1}p_{1}^{n}}{1-p_{1}},\quad n=1,2,\ldots

∎

Let us now designate by $F_{1}^{\ast}:X\rightarrow X_{1},\ F_{2}^{\ast}:X\rightarrow X_{2}$ two other apps that check with $F_{1}$ And $F_{2}$ the conditions

(13)		$\displaystyle\rho_{1}\left(F_{1}^{\ast}\left(u,v\right),F_{1}\left(u,v\right)\right)$	$\displaystyle\leq\delta_{1}$
	$\displaystyle\rho_{2}\left(F_{2}^{\ast}\left(u,v\right),F_{2}\left(u,v\right)\right)$	$\displaystyle\leq\delta_{2}$

for everything $\left(u,v\right)\in X,$ Or $\delta_{1}>0,\ \delta_{2}>0$ are two given real numbers.

Besides the iterative process ( 2 ) we consider the following iterative process:

(14)		$\displaystyle\xi_{1}^{\left(n+1\right)}$	$\displaystyle=F_{1}^{\ast}\left(\xi_{1}^{\left(n\right)},\xi_{2}^{\left(n% \right)}\right)$
	$\displaystyle\xi_{2}^{\left(n+1\right)}$	$\displaystyle=F_{2}^{\ast}\left(\xi_{1}^{\left(n+1\right)},\xi_{2}^{\left(n% \right)}\right),\quad n=0,1,\ldots,\ \big{(}\xi_{1}^{\left(0\right)},\xi_{2}^{% \left(0\right)}\big{)}\in X.$

In the following we will proceed to the delimitation of errors in case the root $\left(\bar{x}_{1},\bar{x}_{2}\right)$ of the system ( 1 ) is approximated by elements of the sequences $\left(\xi_{1}^{\left(n\right)}\right)_{n=0}^{\infty}\left(\xi_{2}^{\left(n% \right)}\right)_{n=0}^{\infty}$ generated using the method ( 14 ). We obviously have no information concerning the applications $F_{1}^{\ast}$ And $F_{2}^{\ast}$ if not that they verify the relations ( 13 ), this is why we cannot affirm anything relative to the convergence of the sequences $\left(\xi_{1}^{\left(n\right)}\right)_{n=0}^{\infty}$ And $\left(\xi_{2}^{\left(n\right)}\right)_{n=0}^{\infty}.$

We will show later that the process of calculating the elements of sequences $\left(\xi_{1}^{\left(n\right)}\right)_{n=0}^{\infty}$ And $\left(\xi_{2}^{\left(n\right)}\right)_{n=0}^{\infty}$ can certainly be stopped when $\rho_{1}\left(\xi_{1}^{\left(n+1\right)},\xi_{1}^{\left(n\right)}\right)<% \varepsilon_{1}$ And $\rho_{2}\left(\xi_{2}^{\left(n+1\right)},\xi_{2}^{\left(n\right)}\right)<% \varepsilon_{2},$ if $\varepsilon_{1}$ And $\varepsilon_{2}$ are suitably chosen with respect to the numbers $\delta_{1},\delta_{2},a,b,\alpha$ And $\beta.$

We have indeed

	$\displaystyle\rho_{1}\left(\xi_{1}^{\left(n+1\right)},\xi_{1}^{\left(n\right)}% \right)=$	$\displaystyle\rho_{1}\left(F_{1}^{\ast}\left(\xi_{1}^{\left(n\right)},\xi_{2}^% {\left(n\right)}\right),F_{1}^{\ast}\left(\xi_{1}^{\left(n-1\right)},\xi_{2}^{% \left(n-1\right)}\right)\right)$
	$\displaystyle\leq$	$\displaystyle\rho_{1}\left(F_{1}^{\ast}\left(\xi_{1}^{\left(n\right)},\xi_{2}^% {\left(n\right)}\right),F_{1}\left(\xi_{1}^{\left(n\right)},\xi_{2}^{\left(n% \right)}\right)\right)$
		$\displaystyle+\rho_{1}\left(F_{1}\left(\xi_{1}^{\left(n\right)},\xi_{2}^{\left% (n\right)}\right),F_{1}\left(\xi_{1}^{\left(n-1\right)},\xi_{2}^{\left(n-1% \right)}\right)\right)$
		$\displaystyle+\rho_{1}\left(F_{1}\left(\xi_{1}^{\left(n-1\right)},\xi_{2}^{% \left(n-1\right)}\right),F_{1}^{\ast}\left(\xi_{1}^{\left(n-1\right)},\xi_{2}^% {\left(n-1\right)}\right)\right)$
	$\displaystyle\leq$	$\displaystyle\alpha\rho_{1}\left(\xi_{1}^{\left(n\right)},\xi_{1}^{\left(n-1% \right)}\right)+\beta\rho_{2}\left(\xi_{2}^{\left(n\right)},\xi_{2}^{\left(n-1% \right)}\right)+2\delta_{1}.$

For $\rho_{2}\left(\xi_{2}^{\left(n+1\right)},\xi_{2}^{\left(n\right)}\right)$ we have the same way

\rho_{2}\left(\xi_{2}^{\left(n+1\right)},\xi_{2}^{\left(n\right)}\right)\leq a% \rho_{1}\left(\xi_{1}^{\left(n+1\right)},\xi_{1}^{\left(n\right)}\right)+b\rho% _{2}\left(\xi_{2}^{\left(n\right)},\xi_{2}^{\left(n-1\right)}\right)+2\delta_{% 2}.

If we write now $f_{n}^{\ast}=\rho_{1}\left(\xi_{1}^{\left(n+1\right)},\xi_{1}^{\left(n\right)}% \right),g_{n}^{\ast}=\rho_{2}\left(\xi_{2}^{\left(n+1\right)},\xi_{2}^{\left(n% \right)}\right),$ $n=0,1,\ldots$ then the above inequalities are written

(15)		$\displaystyle f_{n}^{\ast}$	$\displaystyle\leq\alpha f_{n-1}^{\ast}+\beta\ g_{n-1}^{\ast}+2\delta_{1}$
	$\displaystyle g_{n}^{\ast}$	$\displaystyle\leq a\ f_{n}^{\ast}+b\ g_{n-1}^{\ast}+2\delta_{2}.$

If we place ourselves in the hypotheses of Theorem 1 , then there exists a constant $C_{1}$ independent of $n$ , such that we have the following relations:

(16)		$\displaystyle f_{n}^{\ast}$	$\displaystyle\leq C_{1}h_{1}^{n-1}k_{1}^{n-1}+\tfrac{2\left[\beta\delta_{2}+% \left(1-b\right)\delta_{1}\right]}{\left(1-\alpha\right)\left(1-b\right)-a\beta}$
	$\displaystyle g_{n}^{\ast}$	$\displaystyle\leq C_{1}h_{1}^{n}k_{1}^{n-1}+\tfrac{2\left[\left(1-\alpha\right% )\delta_{2}+a\delta_{1}\right]}{\left(1-\alpha\right)\left(1-b\right)-a\beta},% \qquad n=1,2,\ldots$

Or $\left(h_{1},k_{1}\right)$ is the solution of the system ( 4 ) which verifies the conditions: $0<h_{1}k_{1}<1,\ h_{1}>0,\ k_{1}>0$ .

Indeed, if we choose $C_{1}$ such as

	$\displaystyle C_{1}\geq\max$	$\displaystyle\left\{\left\|\alpha f_{0}^{\ast}+\beta g_{0}^{\ast}+2\delta_{1}-% \tfrac{2\left[\beta\delta_{2}+\left(1-b\right)\delta_{1}\right]}{\left(1-% \alpha\right)\left(1-b\right)-a\beta}\right\|,\right.$
		$\displaystyle\quad\left.\tfrac{1}{h_{1}}\left\|af_{1}^{\ast}+bg_{0}^{\ast}+2% \delta_{2}-\tfrac{2\left[\left(1-\alpha\right)\delta_{2}+a\delta_{1}\right]}{% \left(1-\alpha\right)\left(1-b\right)-a\beta}\right\|\right\}$

so it is obvious that

	$\displaystyle f_{1}^{\ast}$	$\displaystyle\leq\alpha f_{0}^{\ast}+\beta g_{0}^{\ast}+2\delta_{1}\leq C_{1}+% \tfrac{2\left[\beta\delta_{2}+\left(1-b\right)\delta_{1}\right]}{\left(1-% \alpha\right)\left(1-b\right)-a\beta}$
	$\displaystyle g_{1}^{\ast}$	$\displaystyle\leq a\ f_{1}^{\ast}+b\ g_{0}^{\ast}+2\delta_{2}\leq C_{1}h_{1}+% \tfrac{2\left[\left(1-\alpha\right)\delta_{2}+a\delta_{1}\right]}{\left(1-% \alpha\right)\left(1-b\right)-a\beta}$

and inequalities ( 16 ) are verified for $n=1$ .

We assume that relations ( 16 ) hold for each $n=1,2,\ldots s$ and we will show that they also take place for $n=s+1$ . We deduce from ( 15 )

	$\displaystyle f_{s+1}^{\ast}$	$\displaystyle\leq\alpha C_{1}h_{1}^{s-1}k_{1}^{s-1}+\beta\ C_{1}h_{1}^{s}k_{1}% ^{s-1}+\tfrac{2\alpha\left[\beta\delta_{2}+\left(1-b\right)\delta_{1}\right]}{% \left(1-\alpha\right)\left(1-b\right)-a\beta}+\tfrac{2\beta\left[\left(1-% \alpha\right)\delta_{2}+a\delta_{1}\right]}{\left(1-\alpha\right)\left(1-b% \right)-a\beta}+2\delta_{1}$
		$\displaystyle=C_{1}h_{1}^{s-1}k_{1}^{s-1}\left(\alpha+\beta\ h_{1}\right)+% \tfrac{2\left[\beta\delta_{2}+\left(1-b\right)\delta_{1}\right]}{\left(1-b% \right)\left(1-\alpha\right)-a\beta}$
		$\displaystyle=C_{1}h_{1}^{s}k_{1}^{s}+\tfrac{2\left[\beta\delta_{2}+\left(1-b% \right)\delta_{1}\right]}{\left(1-\alpha\right)\left(1-b\right)-a\beta}.$

We show in the same way that the second inequality ( 16 ) also holds for $n=s+1$ .

If now we assume that

(17)		$\displaystyle\varepsilon_{1}$	$\displaystyle>\tfrac{2\left[\beta\delta_{2}+\left(1-b\right)\delta_{1}\right]}% {\left(1-\alpha\right)\left(1-b\right)-a\beta}$
	$\displaystyle\varepsilon_{2}$	$\displaystyle>\tfrac{2\left[\left(1-\alpha\right)\delta_{2}+a\delta_{1}\right]% }{\left(1-\alpha\right)\left(1-b\right)-a\beta}$

then relations ( 16 ) assure us that there exists a $n^{\prime}\in N$ such as for all $n>n^{\prime}$ we have inequalities $\rho_{1}\left(\xi_{1}^{\left(n+1\right)},\xi_{1}^{\left(n\right)}\right)<% \varepsilon_{1}$ And $\rho_{2}\left(\xi_{2}^{\left(n+1\right)},\xi_{2}^{\left(n\right)}\right)<% \varepsilon_{2}.$ Let us designate by $\varepsilon_{1}$ And $\varepsilon_{2}$ two positive numbers which verify the relations ( 17 ) and $n>n^{\prime}+1$ . We propose to evaluate under these conditions the distances between $\bar{x}_{1}$ And $\xi_{1}^{\left(n+1\right)}$ And $\bar{x}_{2}$ And $\xi_{2}^{\left(n+1\right)},$ that is, the delimitation of errors in the case of solving the system ( 1 ) using an approximation process of the form ( 14 ) where the applications $F_{1}^{\ast}$ And $F_{2}^{\ast}$ depend on $F_{1}$ And $F_{2}$ by the relations ( 13 ). Taking into account the hypotheses in which we have placed ourselves, we will have

	$\displaystyle\rho_{1}\left(\bar{x}_{1},\xi_{1}^{\left(n+1\right)}\right)=$	$\displaystyle\rho_{1}\left(F_{1}\left(\bar{x}_{1},\bar{x}_{2}\right),F_{1}^{% \ast}\left(\xi_{1}^{\left(n\right)},\xi_{2}^{\left(n\right)}\right)\right)$
	$\displaystyle\leq$	$\displaystyle\rho_{1}\left(F_{1}\left(\bar{x}_{1},\bar{x}_{2}\right),F_{1}% \left(\xi_{1}^{\left(n\right)},\xi_{2}^{\left(n\right)}\right)\right)$
		$\displaystyle+\rho_{1}\left(F_{1}\left(\xi_{1}^{\left(n\right)},\xi_{2}^{\left% (n\right)}\right),F_{1}^{\ast}\left(\xi_{1}^{\left(n\right)}l\xi_{2}^{\left(n% \right)}\right)\right)$
	$\displaystyle\leq$	$\displaystyle\alpha\rho_{1}\left(\bar{x}_{1},\xi_{1}^{\left(n\right)}\right)+% \beta\rho_{2}\left(\bar{x}_{2},\xi_{2}^{\left(n\right)}\right)+\delta_{1}.$

We have similarly for $\rho_{2}\left(\bar{x}_{2},\xi_{2}^{\left(n+1\right)}\right):$

\rho_{2}\left(\bar{x}_{2},\xi_{2}^{\left(n+1\right)}\right)\leq a\ \rho_{1}% \left(\bar{x}_{1},\xi_{1}^{\left(n+1\right)}\right)+b\ \rho_{2}\left(\bar{x}_{% 2},\xi_{2}^{\left(n\right)}\right)+\delta_{2}

We deduce from the above inequalities

	$\displaystyle\left(1-\alpha\right)\rho_{1}\left(\bar{x}_{1},\xi_{1}^{\left(n+1% \right)}\right)$	$\displaystyle\leq\alpha\varepsilon_{1}+\beta\rho_{2}\left(\bar{x}_{2},\xi_{2}^% {\left(n\right)}\right)+\delta_{1}$
	$\displaystyle\left(1-b\right)\rho_{2}\left(\bar{x}_{2},\xi_{2}^{\left(n+1% \right)}\right)$	$\displaystyle\leq a\ \rho_{1}\left(\bar{x}_{1},\xi_{1}^{\left(n+1\right)}% \right)+b\ \varepsilon_{2}+\delta_{2}$

that's to say

(18)		$\displaystyle\rho_{1}\left(\bar{x}_{1},\xi_{1}^{\left(n+1\right)}\right)$	$\displaystyle\leq\left(\tfrac{\alpha}{1-\alpha}\varepsilon_{1}+\tfrac{\beta}{1% -\alpha}\rho_{2}\left(\bar{x}_{1},\xi_{2}^{\left(n\right)}\right)+\tfrac{% \delta_{1}}{1-\alpha}\right)$
	$\displaystyle\rho_{2}\left(\bar{x}_{2},\xi_{2}^{\left(n+1\right)}\right)$	$\displaystyle\leq\tfrac{a}{1-b}\rho_{1}\left(\bar{x}_{1},\xi_{1}^{\left(n+1% \right)}\right)+\tfrac{b}{1-b}\varepsilon_{2}+\tfrac{\delta_{2}}{1-b}$

from which it results

(19)

\rho_{2}\left(\bar{x}_{2},\xi_{2}^{\left(n+1\right)}\right)\leq\tfrac{a\left(% \alpha\varepsilon_{1}+\beta\varepsilon_{2}\right)+b\left(1-\alpha\right)% \varepsilon_{2}}{\left(1-\alpha\right)\left(1-b\right)-a\beta}+\tfrac{% \varepsilon_{2}}{2}

Because $n>n^{\prime}+1$ , it follows that the second inequality of ( 18 ) also holds in the case where we put $n$ instead of $n+1,$ that's to say

\rho_{2}\left(\bar{x}_{2},\xi_{2}^{\left(n\right)}\right)\leq\tfrac{a}{1-b}% \rho_{1}\left(\bar{x}_{1},\xi_{1}^{\left(n\right)}\right)+\tfrac{b\varepsilon_% {2}}{1-b}+\tfrac{\delta_{2}}{1-b}

This inequality and the first inequality ( 18 ) give us

(20)

\rho_{1}\left(\bar{x}_{1},\xi_{1}^{\left(n+1\right)}\right)\leq\tfrac{\beta% \left(a\varepsilon_{1}+b\varepsilon_{2}\right)+\alpha\varepsilon_{1}\left(1-b% \right)}{\left(1-\alpha\right)\left(1-b\right)-a\beta}+\tfrac{\varepsilon_{1}}% {2}

Bibliography

[1] ^†^†margin: $\leftarrow$ clickable Păvăloiu, I., The resolution of operational systems using iterative methods, Mathematica, 11(34), 1969, 137–141.
[2] Urabe, M., Error estimation in numerical solution of equation by iteration processes , J. Sci. Hiroshima Univ., Ser. AI, 26 (1962), 77–91.

Error estimations in the numerical solving of systems of equations in metric spaces

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Error estimations in the numerical solving of systems of equations in metric spaces

THEOREM 1 .

Demonstration.

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