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 $\lim_{n \to \infty} \| e_n \|_1 = 0$.