Convergence of The F.P. Method: Solving Systems of Two Nonlin. Eqs.

# Convergence of The Fixed Point Method for Solving Systems of Two Nonlinear Equations

We will now develop criterion to ensure that the successive iterations converge to $(\alpha, \beta)$.

If $(\alpha, \beta)$ is a solution to the system prescribed above, then we have that $\alpha = \phi (\alpha, \beta)$ and $\beta = \psi (\alpha, \beta)$. The solution $(\alpha, \beta)$ is contained in some rectangle, say $D = [a, b] \times [c, d]$, that is, $(\alpha, \beta) \in D$. Suppose that $\phi (x, y)$ and $\psi (x, y)$ both have continuous partial derivatives in $D$. We will have that also:

(1)
\begin{align} \quad \alpha - x_{n+1} = \alpha - \phi(x_n, y_n) = \phi (\alpha, \beta) - \phi(x_n, y_n) \\ \quad \beta - y_{n+1} = \beta - \psi (x_n, y_n) = \psi (\alpha, \beta) - \psi (x_n, y_n) \end{align}

In applying a mean value theorem for multivariable functions, we have that for some point $(\xi_n^*, \eta_n^*)$ and $(\xi_{n*}, \eta_{n*})$ between $(\alpha, \beta)$ and $(x_n, y_n)$ that:

(2)
\begin{align} \quad \alpha - x_{n+1} = ... = \phi (\alpha, \beta) - \phi (x_n, y_n) = (\alpha - x_n) \phi_x (\xi_n^*, \eta_n^*) + (\beta - y_n) \phi_y (\xi_n^*, \eta_n^*) \\ \quad \beta - y_{n+1} = ... = \psi (\alpha, \beta) - \psi (x_n, y_n) = (\alpha - x_n) \psi_x (\xi_{n*}, \eta_{n*}) + (\beta - y_n) \psi_y (\xi_{n*}, \eta_{n*}) \end{align}

Taking the absolute value of both sides for both of the equations above and applying the triangle inequality to the righthand side and we have that:

(3)
\begin{align} \quad \mid \alpha - x_{n+1} \mid \: ≤ \: \mid \alpha - x_n \mid \mid \phi(\xi_n^*, \eta_n^*) \mid + \mid \beta - y_n \mid \mid \phi_y (\xi_n^*, \eta_n^*) \mid \\ \quad \mid \beta - y_{n+1} \mid \: ≤ \: \mid \alpha - x_n \mid \mid \psi(\xi_{n*}, \eta_{n*}) \mid + \mid \beta - y_n \mid \mid \psi_y (\xi_{n*}, \eta_{n*}) \mid \end{align}

Summing the inequalities above and we have that:

(4)
\begin{align} \quad \quad \mid \alpha - x_{n+1} \mid + \mid \beta - y_{n+1} \mid \: ≤ \: \mid \alpha - x_n \mid \left ( \mid \phi_x (\xi_n^*, \eta_n^*) \mid + \mid \psi_x (\xi_{n*}, \eta_{n*}) \mid \right ) + \mid \beta - y_n \mid \left ( \mid \phi_y (\xi_n^*, y_n^*) \mid + \mid \psi_y (\xi_{n*}, \eta_{n*}) \mid \right ) \end{align}

Suppose that for $(x, y) \in D = [a, b] \times [c, d]$ we have that:

(5)
\begin{align} \quad \mid \phi_x (x, y) \mid + \mid \psi_x (x, y) \mid ≤ M < 1 \\ \quad \mid \phi_y (x, y) \mid + \mid \psi_y (x, y) \mid ≤ M < 1 \\ \end{align}

Then this implies that:

(6)
\begin{align} \quad \mid \alpha - x_{n+1} \mid + \mid \beta - y_{n+1} \mid \: ≤ \: M \mid \alpha - x_n \mid + \mid \beta - y_n \mid) \end{align}

Thus if the initial approximation $(x_0, y_0) \in D$ then the successive approximations $(x_n, y_n) \in D$ for $n = 0, 1, 2, ...$. We will use the $1$-norm (defined as $\| x \|_1 = \sum_{i=1}^{n} \mid x_i \mid$) for our error. Let $e_{n+1} = \begin{bmatrix} \alpha - x_{n+1}\\ \beta - y_{n+1} \end{bmatrix}$. We have that:

(7)
\begin{align} \quad \| e_{n+1} \|_1 = \mid \alpha - x_{n+1} \mid + \mid \beta - y_{n+1} \mid \end{align}

Inductively we see that $\| e_{n+1} \|_1 ≤ M^{n+1} \| e_0 \|_1$:

(8)
\begin{align} \quad \quad \| e_{n+1} \|_1 \: ≤ \: M(\mid \alpha - x_n \mid + \mid \beta - y_n \mid) \: ≤ \: M^2 (\mid \alpha - x_{n-1} \mid + \mid \beta - y_{n-1} \mid ) \: ≤ \: ... \: ≤ \: M^{n+1} (\mid \alpha - x_0 \mid + \mid \beta - y_0 \mid) = M^{n+1} \| e_0 \|_1 \end{align}

Therefore for $e_n = \begin{bmatrix} \alpha - x_n\\ \beta - y_n \end{bmatrix}$ we have that $\| e_n \|_1 = \mid \alpha - x_n \mid + \mid \beta - y_n \mid$ and:

(9)
\begin{align} \quad \| e_n \|_1 ≤ M^n \| e_0 \|_1 \end{align}

Thus if $\mid \phi_x (x, y) \mid + \mid \psi_x (x, y) \mid ≤ M < 1$ and $\mid \phi_y (x, y) \mid + \mid \psi_y (x, y) \mid ≤ M < 1$ for all $(x, y) \in D$, then the following sequence of approximations converge to $(\alpha, \beta)$:

(10)
\begin{align} \quad x_{n+1} = \phi (x_n, y_n) \quad , \quad y_{n+1} = \psi (x_n, y_n) \end{align}

This is because $0 ≤ \| e_n \|_1 ≤ M^n \| e_0 \|_1$, and $\lim_{n \to \infty} 0 = \lim_{n \to \infty} M^n \| e_0 \|_1$, so by the squeeze theorem we have that $\lim_{n \to \infty} \| e_n \|_1 = 0$.