"Babeş-Bolyai" University

Faculty of Mathematics and Physics

Research Seminars

Seminar on Mathematical Analysis

Preprint Nr.7, 1988, pp.167-178

Error estimations in the numerical solving of the systems of equations

Ion Pavaloiu

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 will denote by $F_{1}:X\rightarrow X_{1}$ And $F_{2}:X\rightarrow X_{2}$ two applications of space $X$ in $X_{1}$ , respectively in $X_{2}$ and we consider the following system of equations:

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

To solve the system ( 1 ) we will adopt the following Gauss-Seidel type iterative process:

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

In works [ 2 ] , [ 3 ] we studied the convergence of method ( 2 ) under the assumption that the applications $F_{1}$ And $F_{2}$ satisfy Lipschitz-type conditions throughout the space $X$ .

In work [ 4 ] we obtained bounds of errors in the numerical resolution of system ( 1 ) using a method of type ( 2 ), in which the applications $F_{1}\$ And $F_{2}$ are replaced by two other applications $F_{1}^{\ast}$ And $F_{2}^{\ast}$ , which meet certain conditions for bringing together $F_{2}$ And $F_{1}$ throughout space $X$ .

In the applications of this theory to the resolution of a class of concrete equations, the conditions imposed on the applications $F_{1},F_{2},F_{1}^{\ast}$ And $F_{2}^{\ast}$ throughout space $X$ are awkward, because of the fact that the space $X$ may not be limited.

In the following we will study this problem on the assumption that $F_{1}$ And $F_{2}$ meet Lipschitz-type conditions and rapprochement conditions $F_{1}^{\ast}$ And $F_{2}^{\ast}$ in certain bounded subsets $D_{1}\subset X_{1}$ And $D_{2}\subset X_{2}$ .

Let us therefore designate by $D_{1}$ And $D_{2}$ two bounded sets of spaces $X_{1}$ respectively $X_{2}$ and by $D=D_{1}\times D_{2}$ their Cartesian product. Consider two sequences $\left(f_{n}\right)_{n=0}^{\infty}$ And $\left(g_{n}\right)_{n=0}^{\infty}$ whose elements meet the conditions:

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

Or $\alpha,\beta,$ $a$ And $b$ are non-negative real numbers and $f_{n}\geq 0,$ $g_{n}\geq 0$ for everything $n=0,1,\ldots$ .

We will associate to the system ( 3 ) the following system of equations in the unknowns $h$ And $k$ :

(4)		$\displaystyle\alpha+\beta h$	$\displaystyle=h\ k$
	$\displaystyle a\ k+b$	$\displaystyle=n\ k$

In works [ 2 ] , [ 3 ] and [ 4 ] we have shown that if the numbers $\alpha,\beta,\ a$ And $b$ check the relationships

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

then the system ( 4 ) admits real solutions $\left(h_{i},k_{i}\right),\ i=1,2$ for which $0<h_{i}k_{i}<1$ and that one of the solutions verifies the conditions $h_{1}>0,\ k_{1}>0$ . In addition, the elements of the suites $\left(f_{n}\right)_{n=0}^{\infty}$ $\left(g_{n}\right)_{n=0}^{\infty}$ check the relationships

(6)		$\displaystyle f_{n}$	$\displaystyle\leq C\ h_{1}^{n-1}\ k_{1}^{n-1}$
	$\displaystyle g_{n}$	$\displaystyle\leq C\ h_{1}^{n}\ k_{1}^{n-1},\qquad n=1,2,\ldots,$

Or $C=\max\{\alpha f_{0}+\beta g_{0},\frac{af_{1}+bg_{0}}{h_{1}}\}.$

If we write $p_{1}=h_{1},k_{1}$ then we easily see using ( 4 ) that $p_{1}$ check the equation

(7)

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

Let us designate by $d_{1}>0$ a number such that $S_{1}\subseteq D_{1}$ , $S_{2}\subseteq D_{2}$ , Or

(8)		$\displaystyle S_{1}$	$\displaystyle=\left\{x\in X_{1}:\rho_{1}\big{(}x,x_{1}^{\left(0\right)}\big{)}% \leq\tfrac{d_{1}}{1-p_{1}}\right\}$
	$\displaystyle S_{2}$	$\displaystyle=\left\{x\in X_{2}:\rho_{2}\big{(}x,x_{2}^{\left(0\right)}\big{)}% \leq\tfrac{d_{1}h_{1}}{1-p_{1}}\right\}$

Regarding the convergence of sequences $\big{(}x_{1}^{\left(n\right)}\big{)}_{n=0}^{\infty},\ \big{(}x_{2}^{\left(n% \right)}\big{)}_{n=0}^{\infty}$ we have the following theorem:

Theorem 1 .

If the applications $F_{1},F_{2}$ check the conditions

i.

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

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

for all $\left(x_{1},x_{2}\right),\ \left(y_{1},y_{2}\right)\in D;$
ii.

The names $\alpha,\ \beta,$ $a$ And $b$ check the relationships ( 5 );
iii.

The elements $x_{1}^{\left(1\right)}$ And $x_{2}^{\left(1\right)}$ suites $\big{(}x_{1}^{\left(n\right)}\big{)}_{n=0}^{\infty},\big{(}x_{2}^{\left(n% \right)}\big{)}_{n=0}^{\infty}$ check the conditions $\rho_{1}\big{(}x_{1}^{\left(0\right)},x_{1}^{\left(1\right)}\big{)}<d_{1}$ And $\rho_{2}\big{(}x_{2}^{\left(0\right)},x_{2}^{\left(1\right)}\big{)}\leq d_{1}h% _{1}$ .

Then the following properties are true:

j.

These suites $\big{(}x_{1}^{\left(n\right)}\big{)}_{n=0}^{\infty},\big{(}x_{2}^{\left(n% \right)}\big{)}_{n=0}^{\infty}$ constructed using method ( 2 ) are convergent;
jj.

If we write $\bar{x}_{1}=\lim\limits_{n\rightarrow\infty}x_{1}^{\left(n\right)}$ And $\bar{x}_{2}=\lim\limits_{n\rightarrow\infty}x_{2}^{\left(n\right)}$ SO $\left(\bar{x}_{1},\bar{x}_{2}\right)\in S$ Or $S=S_{1}\times S_{2}$ And $\left(\bar{x}_{1},\bar{x}_{2}\right)$ is the only solution of the system ( 1 ) in the set $S$ ;
jjj.

The following relationships take place

$\displaystyle\rho_{1}\left(\bar{x}_{1},x_{1}^{\left(n\right)}\right)$ $\displaystyle\leq\tfrac{d_{1}p_{1}^{n}}{1-p_{1}}$

$\displaystyle\rho_{2}\left(\bar{x}_{2},x_{2}^{\left(n\right)}\right)$ $\displaystyle\leq\tfrac{d_{1}h_{1}p_{1}^{n}}{1-p_{1}}$

Demonstration.

From i. and ( 2 ) we deduce the following inequalities

	$\displaystyle\rho_{1}\left(x_{1}^{\left(2\right)},x_{1}^{\left(1\right)}\right)$	$\displaystyle\leq\alpha\rho_{1}\left(x_{1}^{\left(1\right)},x_{1}^{\left(0% \right)}\right)+\beta\rho_{2}\left(x_{1}^{\left(1\right)},x_{2}^{\left(0\right% )}\right)$
		$\displaystyle\leq\alpha d_{1}+\beta d_{1}h_{1}\leq d_{1}p_{1}$

And

	$\displaystyle\rho_{2}\left(x_{2}^{\left(2\right)},x_{2}^{\left(1\right)}\right)$	$\displaystyle\leq a\rho_{1}\left(x_{1}^{\left(2\right)},x_{1}^{\left(1\right)}% \right)+b\rho_{2}\left(x_{2}^{\left(1\right)},x_{2}^{\left(0\right)}\right)$
		$\displaystyle\leq ad_{1}p_{1}+bd_{1}h_{1}\leq d_{1}h_{1}p_{1}.$

We now show that $x_{1}^{\left(2\right)}\in S_{1}$ And $x_{2}^{\left(2\right)}\in S_{2}$ . We have indeed

	$\displaystyle\rho_{1}\left(x_{1}^{\left(2\right)},x_{1}^{\left(0\right)}\right)$	$\displaystyle\leq\rho_{1}\left(x_{1}^{\left(2\right)},x_{1}^{\left(1\right)}% \right)+\rho_{1}\left(x_{1}^{\left(1\right)},x_{1}^{\left(0\right)}\right)$
	$\displaystyle d_{1}+d_{1}\ p_{1}$	$\displaystyle<\tfrac{d_{1}}{1-p_{1}},$

And

	$\displaystyle\rho_{2}\left(x_{2}^{\left(2\right)},x_{2}^{\left(0\right)}\right)$	$\displaystyle\leq\rho_{2}\left(x_{2}^{\left(2\right)},x_{2}^{\left(1\right)}% \right)+\rho_{2}\left(x_{2}^{\left(1\right)},x_{2}^{\left(0\right)}\right)$
		$\displaystyle\leq d_{1}h_{1}+d_{1}\ p_{1}h_{1}\leq\tfrac{h_{1}d_{1}}{1-p_{1}}$

from which it follows that $x_{1}^{\left(2\right)}\in S_{1},x_{2}^{\left(2\right)}\in S_{2}$ .

We will now assume that the following inequalities hold

(9)		$\displaystyle\rho_{1}\left(x_{1}^{\left(i\right)},x_{1}^{\left(i-1\right)}\right)$	$\displaystyle\leq d_{1}p_{1}^{i-1}$
	$\displaystyle\rho_{2}\left(x_{2}^{\left(i\right)},x_{2}^{\left(i-1\right)}\right)$	$\displaystyle\leq d_{1}h_{1}p_{1}^{i-1}$

For $i=1,2,\ldots,k$ And

(10)

x_{1}^{\left(k\right)}\in S_{1},\ x_{2}^{\left(k\right)}\in S_{2}.

Taking into account ii., we deduce from hypotheses ( 9 ) and ( 10 ) and from i.

	$\displaystyle\rho_{1}\left(x_{1}^{\left(k+1\right)},x_{1}^{\left(k\right)}\right)$	$\displaystyle=\rho_{1}\left(F_{1}(x_{1}^{\left(k\right)},x_{2}^{\left(k\right)% }),F_{1}(x_{1}^{\left(k-1\right)},x_{2}^{\left(k-1\right)})\right)$
		$\displaystyle\leq\alpha\rho_{1}\left(x_{1}^{\left(k\right)},x_{1}^{\left(k-1% \right)}\right)+\beta\rho_{2}\left(x_{2}^{\left(k\right)},x_{2}^{\left(k-1% \right)}\right)$
		$\displaystyle\leq d_{1}p_{1}^{k-1}\left(\alpha+\beta h_{1}\right)=d_{1}p_{1}^{k}$

and taking into account the above inequality we deduce in an analogous manner

\rho_{2}\left(x_{2}^{\left(k+1\right)},x_{2}^{\left(k\right)}\right)\leq d_{1}% \cdot h_{1}\cdot p_{1}^{k}

from which it follows that relations ( 9 ) also hold for $i=k+1$ . We will now show that $x_{1}^{\left(k+1\right)}\in S_{1}$ And $x_{2}^{\left(k+1\right)}\in S_{2}$ . It follows from the above inequalities

	$\displaystyle\rho_{1}\left(x_{1}^{\left(k+1\right)},x_{1}^{\left(0\right)}\right)$	$\displaystyle\leq\rho_{1}\left(x_{1}^{\left(k+1\right)},x_{1}^{\left(k\right)}% \right)+\rho_{1}\left(x_{1}^{\left(k\right)},x_{1}^{\left(k-1\right)}\right)+% \ldots\rho_{1}\left(x_{1}^{\left(1\right)},x_{1}^{\left(0\right)}\right)$
		$\displaystyle\leq d_{1}+d_{1}p_{1}+d_{1}p_{1}^{2}+\ldots+d_{1}p_{1}^{k}<\tfrac% {d_{1}}{1-p_{1}}$

and similarly

\rho_{2}\left(x_{2}^{\left(k+1\right)},x_{2}^{\left(0\right)}\right)<\tfrac{d_% {1}h_{1}}{1-p_{1}}.

Since inequalities ( 9 ) hold for all $i=1,2,\ldots,$ it results that the suites $\left(x_{1}^{\left(n\right)}\right)_{n=0}^{\infty}$ And $\left(x_{2}^{\left(n\right)}\right)_{n=0}^{\infty}$ are fundamental and that the following inequalities are true

(11)		$\displaystyle\rho_{1}\left(x_{1}^{\left(n+s\right)},x_{1}^{\left(n\right)}\right)$	$\displaystyle\leq\tfrac{d_{1}p_{1}^{n}}{1-p_{1}}$
	$\displaystyle\rho_{2}\left(x_{2}^{\left(n+s\right)},x_{2}^{\left(n\right)}\right)$	$\displaystyle\leq\tfrac{d_{1}h_{1}p_{1}^{n}}{1-p_{1}}$

for everything $n=0,1,\ldots;$ $s=1,2,\ldots$ .

From the hypothesis that the spaces $X_{1}$ And $X_{2}$ are complete spaces it follows that the sequences $\left(x_{n}^{\left(n\right)}\right)_{n=0}^{\infty}$ And $\left(x_{2}^{\left(n\right)}\right)_{n=0}^{\infty}$ are convergent.

Passing the limit for $s\rightarrow\infty$ we deduce from ( 11 )

(12)		$\displaystyle\rho_{1}\left(\bar{x}_{1},x_{1}^{\left(n\right)}\right)$	$\displaystyle\leq\tfrac{d_{1}p_{1}^{n}}{1-p_{1}}$
	$\displaystyle\rho_{2}\left(\bar{x}_{2},x_{2}^{\left(n\right)}\right)$	$\displaystyle\leq\tfrac{d_{1}h_{1}p_{1}^{n}}{1-p_{1}}$

It results, for $n=0,$ what $\bar{x}_{1}\in S_{1}$ And $\bar{x}_{2}\in S_{2}$ . We will show that $\left(\bar{x}_{1},\bar{x}_{2}\right)$ is the only solution of the system ( 1 ) to the property that $\bar{x}_{1}\in S_{1}$ And $\bar{x}_{2}\in S_{2}$ .

Let us suppose by contradiction that system ( 1 ) has another solution $\left(\bar{y}_{1},\bar{y}_{2}\right)$ to the property $\bar{y}_{1}\in S_{1}$ , $\bar{y}_{2}\in S_{2}$ . Taking into account i., the inequalities result

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

and from ( 5 ) it follows that $\frac{\beta a}{\left(1-\alpha\right)\left(1-b\right)}<1$ . Therefore the above inequalities hold if and only if $\bar{x}_{1}=\bar{y}_{1}$ And $\bar{x}_{2}=\bar{y}_{2}$ . ∎

We now consider two applications $F_{1}^{\ast}:D\rightarrow X_{1}$ And $F_{2}^{\ast}:D\rightarrow X_{2}$ , Or $D=D_{1}\times D_{2}$ .

We will assume that $F_{1}^{\ast},$ $F_{2}^{\ast},$ $F_{1}$ And $F_{2}$ check the relationships

(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 $\left(u,v\right)\in D,$ Or $\delta_{1}>0,$ $\delta_{2}>0$ are given numbers. In order to solve the system ( 1 ) we will now consider instead of the procedure ( 2 ) the following iterative procedure:

(14)		$\displaystyle\xi_{1}^{\left(n+1\right)}$	$\displaystyle=F_{1}\left(\xi_{1}^{\left(n\right)},\xi_{2}^{\left(n\right)}\right)$
	$\displaystyle\xi_{2}^{\left(n+1\right)}$	$\displaystyle=F_{2}\left(\xi_{1}^{\left(n+1\right)},\xi_{2}^{\left(n\right)}% \right),\quad n=0,1,\ldots\xi_{1}^{\left(0\right)}=x_{1}^{\left(0\right)},\xi_% {2}^{\left(0\right)}=x_{2}^{\left(0\right)},$

In the following we will proceed to the delimitation of errors in the case where for the resolution of the system ( 1 ) we use instead of the method ( 2 ) the method ( 14 ). We write

(15)		$\displaystyle\theta_{1}$	$\displaystyle=\tfrac{\beta\delta_{2}+\left(1-b\right)\delta_{1}}{\left(1-% \alpha\right)\left(1-b\right)-a\beta}$
	$\displaystyle\theta_{2}$	$\displaystyle=\tfrac{\left(1-\alpha\right)\delta_{2}+a\delta_{1}}{\left(1-% \alpha\right)\left(1-b\right)-a\beta}$

and designate by $S_{1}^{\ast},S_{2}^{\ast}$ the following sets:

(16)		$\displaystyle S_{1}^{\ast}$	$\displaystyle=\left\{x\in X_{1}:\rho_{1}\left(x,x_{1}^{\left(0\right)}\right)% \leq d_{1}+\tfrac{d_{1}}{1-p_{1}}+\theta_{1}\right\}$
	$\displaystyle S_{2}^{\ast}$	$\displaystyle=\left\{x\in X_{2}:\rho_{2}\left(x,x_{2}^{\left(0\right)}\right)% \leq d_{1}h_{1}+\tfrac{d_{1}h_{1}}{1-p_{1}}+\theta_{2}\right\}.$

In these notations we will demonstrate the following theorem:

Theorem 2 .

If the assumptions of Theorem 1 are satisfied and if in addition the following conditions are met:

i ${}_{1}.$

The applications $F_{1},F_{2},F_{1}^{\ast}$ And $F_{2}^{\ast}$ meet the conditions ( 13 );
i ${}_{2}.$

$S_{1}^{\ast}\subseteq D_{1},\ S_{2}^{\ast}\subseteq D_{2}$ ,

so whatever the real numbers are $\varepsilon_{1},$ $\varepsilon_{2}$ who verify the relationships $\varepsilon_{1}>2\theta_{1},\ \varepsilon_{2}>2\theta_{2}$ , there is a $n^{\prime}\in{\mathbb{N}}$ such as $\rho_{1}\left(\xi_{1}^{\left(n+1\right)},\xi_{1}^{\left(n\right)}\right)\leq% \varepsilon_{1}$ , And $\rho_{2}\left(\xi_{2}^{\left(n+1\right)},\xi_{2}^{\left(n\right)}\right)\leq% \varepsilon_{2}$ for everything $n>n^{\prime}$ and, furthermore, the following properties are true:

j ${}_{1}.$

$\xi_{1}^{\left(n\right)}\in S_{1},$ $\xi_{2}^{\left(n\right)}\in S_{2}$ for everything $n=1,2,\ldots;$
j ${}_{2}.$

The following inequalities are true:

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

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

for everything $n>n^{\prime}$ , Or $\left(\bar{x}_{1},\bar{x}_{2}\right)$ is the solution of the system ( 1 ).

Demonstration.

We will first show that properties j ₁ are true. Since we have assumed that $\xi_{1}^{\left(0\right)}=x_{1}^{\left(0\right)},\xi_{2}^{\left(0\right)}=x_{2}% ^{\left(0\right)},$ it follows from ( 14 ) and i.

	$\displaystyle\rho_{1}\left(x_{1}^{\left(1\right)},\xi_{1}^{\left(1\right)}\right)$	$\displaystyle=\rho_{1}\left(F_{1}\left(x_{1}^{\left(0\right)},x_{2}^{\left(0% \right)}\right),F_{1}^{\ast}\left(\xi_{1}^{\left(0\right)},\xi_{2}^{\left(0% \right)}\right)\right)$
		$\displaystyle\leq\delta_{1}+\alpha\rho_{1}\left(\xi_{1}^{\left(0\right)}x_{1}^% {\left(0\right)}\right)+\beta\rho_{2}\left(\xi_{2}^{\left(0\right)}x_{2}^{% \left(0\right)}\right)=\delta_{1}$

Taking into account iii., we deduce

	$\displaystyle\rho_{1}\left(\xi_{1}^{\left(1\right)},x_{1}^{\left(0\right)}\right)$	$\displaystyle\leq\rho_{1}\left(\xi_{1}^{\left(1\right)},x_{1}^{\left(1\right)}% \right)+\rho_{1}\left(x_{1}^{\left(1\right)},x_{1}^{\left(0\right)}\right)$
		$\displaystyle\leq\delta_{1}+d_{1}\leq d_{1}+\tfrac{d_{1}}{1-p_{1}}+\theta_{1}$

since obviously $\delta_{1}\leq\theta_{1}$ , from which it results $\xi_{1}^{\left(1\right)}\in S_{1}^{\ast}$ .

We have in a similar way

	$\displaystyle\rho_{2}\left(x_{2}^{\left(1\right)},\xi_{2}^{\left(1\right)}\right)$	$\displaystyle=\rho_{2}\left(F_{2}\left(x_{1}^{\left(1\right)},x_{2}^{\left(0% \right)}\right),F_{2}^{\ast}\left(\xi_{1}^{\left(1\right)},\xi_{2}^{\left(0% \right)}\right)\right)$
		$\displaystyle\leq\delta_{2}+a\rho_{1}\left(\xi_{1}^{\left(1\right)},x_{1}^{% \left(1\right)}\right)+b\rho_{2}\left(\xi_{2}^{\left(0\right)},x_{2}^{\left(0% \right)}\right)\leq\delta_{2}+a\delta_{1}$

Taking into account iii. we deduce

\rho_{2}\left(\xi_{2}^{\left(1\right)},x_{2}^{\left(0\right)}\right)\leq\rho_{% 2}\left(\xi_{2}^{\left(1\right)},x_{2}^{\left(1\right)}\right)+\rho_{2}\left(x% _{2}^{\left(1\right)},x_{2}^{\left(0\right)}\right)\leq\delta_{2}+a\delta_{1}+% d_{1}h_{1}.

Taking into account that $\delta_{2}+a\delta_{1}\leq\theta_{2}$ it follows that

\rho_{2}\left(\xi_{2}^{\left(1\right)},x_{2}^{\left(0\right)}\right)\leq h_{1}% d_{1}+\tfrac{h_{1}d_{1}}{1-p_{1}}+\theta_{2}

and therefore $\xi_{2}^{\left(1\right)}\in S_{2}^{\ast}$ .

We will now assume by induction that $\xi_{1}^{\left(n-1\right)}\in S_{1}^{\ast}$ And $\xi_{2}^{\left(n-1\right)}\in S_{2}^{\ast}$ .

We then have

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

And

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

We show by induction, from the above relations, that

(19)		$\displaystyle\rho_{1}\left(\xi_{1}^{\left(n\right)},x_{1}^{\left(n\right)}\right)$	$\displaystyle\leq d_{1}h_{1}^{n-1},k_{1}^{n-1}+\theta_{1}$
	$\displaystyle\rho_{2}\left(\xi_{2}^{\left(n\right)},x_{2}^{\left(n\right)}\right)$	$\displaystyle\leq d_{1}h_{1}^{n},k_{1}^{n-1}+\theta_{2},$

from which we deduce:

\rho_{1}\left(\xi_{1}^{\left(n\right)},x_{1}^{\left(0\right)}\right)\leq\rho_{% 1}\left(\xi_{1}^{\left(n\right)},x_{1}^{\left(n\right)}\right)+\rho_{1}\left(x% _{1}^{\left(n\right)},x_{1}^{\left(0\right)}\right)\leq d_{1}+\tfrac{d_{1}}{1-% p_{1}}+\theta_{1}

And

\rho_{2}\left(\xi_{2}^{\left(n\right)},x_{2}^{\left(0\right)}\right)\leq\rho_{% 2}\left(\xi_{2}^{\left(n\right)},x_{2}^{\left(n\right)}\right)+\rho_{2}\left(x% _{2}^{\left(n\right)},x_{2}^{\left(0\right)}\right)\leq d_{1}h_{1}+\tfrac{d_{1% }h_{1}}{1-p_{1}}+\theta_{2}

that's to say $\xi_{1}^{\left(n\right)}\in S_{1}^{\ast},$ $\xi_{2}^{\left(n\right)}\in S_{2}^{\ast}$ .

From the assumptions of Theorem 2 and the fact that $\xi_{1}^{\left(n\right)}\in S_{1}^{\ast},\xi_{2}^{\left(n\right)}\in S_{2}^{\ast}$ for everything $n\in N$ , the following relationships result:

\rho_{1}\left(\xi_{1}^{\left(n+1\right)},\xi_{1}^{\left(n\right)}\right)\leq% \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}

And

\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}

For $n=0,1,\ldots$ .

From the above inequalities we deduce the following inequalities:

(20)		$\displaystyle\rho_{1}\left(\xi_{1}^{\left(n+1\right)},\xi_{1}^{\left(n\right)}\right)$	$\displaystyle\leq d_{1}h_{1}^{n-1}\cdot k_{1}^{n-1}+2\theta_{1}$
	$\displaystyle\rho_{2}\left(\xi_{2}^{\left(n+1\right)},\xi_{2}^{\left(n\right)}\right)$	$\displaystyle\leq d_{1}h_{1}^{n}\cdot k_{1}^{n-1}+2\theta_{2},\qquad n=1,2,\ldots$

We immediately deduce from ( 20 ) that if $\varepsilon_{1}>2\theta_{1},$ And $\varepsilon_{2}>2\theta_{2},$ so it exists $n^{\prime}\in{\mathbb{N}}$ such as for $n>n^{\prime}$ $\rho_{1}\left(\xi_{1}^{\left(n+1\right)},\xi_{1}^{\left(n\right)}\right)\leq% \varepsilon_{1}$ And $\rho_{2}\left(\xi_{2}^{\left(n+1\right)},\xi_{2}^{\left(n\right)}\right)\leq% \varepsilon_{2}$ that is, the iterative process ( 14 ) can be stopped when the distance between two successive iterations is sufficiently small.

We will now evaluate the distances between the solutions $\bar{x}_{1}$ And $\bar{x}_{2}$ of the system ( 1 ) and the elements of the sequences $\left(\xi_{1}^{\left(n\right)}\right)_{s=0}^{\infty},$ respectively $\left(\xi_{2}^{\left(n\right)}\right)_{s=0}^{\infty}$ . We will assume that the iterative process ( 14 ) is stopped when $\rho_{1}\left(\xi_{1}^{\left(n+1\right)},\xi_{1}^{\left(n\right)}\right)\leq% \varepsilon_{1}$ And $\rho_{2}\left(\xi_{2}^{\left(n+1\right)},\xi_{2}^{\left(n\right)}\right)\leq% \varepsilon_{2}$ .

Taking into account what has been demonstrated above and the assumptions of Theorem 2 , we have:

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

It follows 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

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

from which it results

(22)

\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 ( 21 ) is also true if we replace $n+1$ about $n$ , that is to say that

\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}

inequality which associated with the inequality of ( 21 ) gives us

(23)

\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\left(1-b\right)% \varepsilon_{1}}{\left(1-\alpha\right)\left(1-b\right)-a\beta}+\tfrac{% \varepsilon_{1}}{2}

∎

Bibliography

[1]
[2] Pavaloiu, I., ^†^†margin: $\leftarrow$ clickable
Introduction to the theory of approximation of solutions of equations , Dacia Publishing House, Cluj-Napoca, 1976.
[3] ^†^†margin: $\leftarrow$ clickable
$\leftarrow$ clickable
$\leftarrow$ clickable Păvăloiu, I., The resolution of operational systems using iterative methods , Mathematica, 11(34), (1969), 137–141.
[4] Păvăloiu, I., Estimation of errors in the numerical resolution of systems of equations in metric spaces , Seminar on Functional Analysis and Numerical Methods, Preprint Nr. 1, (1987), 121–129.
[5] Păvăloiu, I., The convergence of certain iterative methods for solving certain operatorial equations , Seminar on Functional analysis and Numerical Methods, Preprint Nr. 1 (1986), 127–132.
[6] Traub, J. F., Iterative Methods for the Solution of Equations, Prentice Hall Series in Automatic Computation, Englewood Cliffs, N. J. (1964).
[7] Urabe, M., Convergence of numerical iteration in solution of equations, J. Sci. Hiroshima Univ. Ser. A, 19 (1956), 479–489.
[8] Urabe, M., Error estimation in numerical solution of equations by iteration process, J. Sci. Hiroshima Univ. Ser. A-I, 26, (1962), 77–91.

Résumé

In this work, bounds are provided for the errors committed in the numerical resolution using the Gauss-Seidel method - of a system of two equations with two unknowns in metric spaces.

And $\left(X_{i},\rho_{i}\right),\ i=1,2,$ are two complete metric spaces and $F_{1}:X_{1}\times X_{2}\rightarrow X_{1}$ And $F_{2}:X_{1}\times X_{2}\rightarrow X_{2}$ two applications, then we apply for the resolution of the system $x_{1}=F_{1}\left(x_{1},x_{2}\right),\ x_{2}=F_{2}\left(x_{1},x_{2}\right),$ the Gauss-Seidel method and sufficient conditions are given for the convergence of the process used.

We designate by $D_{1}\subset X_{1}$ And $D_{2}\subset X_{2}$ two bounded sets of $X_{1}$ And $X_{2}$ and then we consider two operators $F_{1}^{\ast}:D_{1}\times D_{2}\rightarrow X_{1}\ F_{2}^{\ast}:D_{1}\times D_{2% }\rightarrow X_{2}$ which check against $F_{1}$ And $F_{2}$ the conditions: $\rho_{1}\left(F_{1}\left(x_{1},x_{2}\right),F_{1}^{\ast}\left(x_{1},x_{2}% \right)\right)\leq\varepsilon_{1},\ \rho_{2}\left(F_{2}\left(x_{1},x_{2}\right% ),F_{2}^{\ast}\left(x_{1},x_{2}\right)\right)\leq\varepsilon_{2}$ for each $\left(x_{1},x_{2}\right)\in D_{1}\times D_{2}$ . Under these conditions, the Gauss-Seidel method is applied to the system for the resolution of the initial system. $x_{1}=F_{1}^{\ast}\left(x_{1},x_{2}\right),\ x_{2}=F_{2}^{\ast}\left(x_{1},x_{% 2}\right)$ . Under these conditions, we give delimitations of the distance between the solution of the system and the approximate solution.

Ion Pavaloiu

Institute of Mathematics

Post Office 1

C.P. 68

3400 Cluj-Napoca

Romania

This Note is in final form and no version of it is or will be submitted for publication elsewhere.

Error estimations in the numerical solving of the systems of equations

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Error estimations in the numerical solving of the systems of equations

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