Applying The F.P. Method for Solving Systems of Two Nonlinear Eqs.

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

Be sure to review the following pages regarding The Fixed Point method for solving systems of two nonlinear equations:

We will now look at an example of applying this method.

## Example 1

Consider the following system of nonlinear equations $\left\{\begin{matrix} y = \frac{1}{3} \ln (x + 2)\\ y = e^{3x} - 2 \end{matrix}\right.$. There exists a solution $(\alpha, \beta)$ such that $\alpha, \beta > 0$, and there exists a solution $(\tilde{\alpha}, \tilde{\beta})$ such that $\tilde{\alpha}, \tilde{\beta} < 0$. This system can be rewritten as $\left\{\begin{matrix} x = \phi_1(x, y) = e^{3y} - 2\\ y = \psi_1 (x, y) = e^{3x} - 2 \end{matrix}\right.$ and as $\left\{\begin{matrix} x = \phi_2(x, y) = \frac{1}{3} \ln (y + 2)\\ y = \psi_2 (x, y) = \frac{1}{3} \ln (x + 2) \end{matrix}\right.$. Determine which choice of $\phi$ and $\psi$ are suitable for the convergence of The Fixed Point method for each of the solutions $(\alpha, \beta)$ and $(\tilde{\alpha}, \tilde{\beta})$, and using the initial approximation $(x_0, y_0) = (-1.5, -1.5)$ to compute successive approximations $(x_n, y_n)$ of $(\tilde{\alpha}, \tilde{\beta})$ until $\| (x_n, y_n) - (x_{n-1}, y_{n-1}) \|_1 < \epsilon = 10^{-3}$.

Let's first consider $\left\{\begin{matrix} x = \phi_1(x, y) = e^{3y} - 2\\ y = \psi_1 (x, y) = e^{3x} - 2 \end{matrix}\right.$. We have that $\phi_{1, x} = 0$ AND $\psi_{1, x} = e^{3x}$. We therefore have that $\mid \phi_{1, x} \mid + \mid \psi_{1, x} \mid = 0 + e^{3x} < 1$ for $x < \frac{1}{3}$, i.e, $\mid \phi_{1, x} \mid + \mid \psi_{1, x} \mid < \frac{1}{3}$ for $x < 0$. We also have that $\mid \phi_{1, y} \mid = e^{3y}$ and $\mid \psi_{1, y} = 0$ and so $\mid \phi_{1, y} \mid + \mid \psi_{1, y} \mid = e^{3y} < 1$ for $y < \frac{1}{3}$, i.e, $\mid \phi_{1, y} \mid + \mid \psi_{1, y} \mid < 1$ for $y < 0$. Therefore this choice of rewriting the prescribed system of equations is suitable for the negative solution $(\tilde{\alpha}, \tilde{\beta})$.

Now let's consider $\left\{\begin{matrix} x = \phi_2(x, y) = \frac{1}{3} \ln (y + 2)\\ y = \psi_2 (x, y) = \frac{1}{3} \ln (x + 2) \end{matrix}\right.$. We have that $\phi_{2, x} = 0$ and $\psi_{2, x} = \frac{1}{3(x + 2)}$. Thus $\mid \phi_{2, x} \mid + \mid \psi_{2, x} \mid = \frac{1}{3(x + 2)} < 1$ for $x > 0$. Furthermore, we have that $\phi_{2, y} = \frac{1}{3(y + 2)}$ and $\psi_{2, y} = 0$ and so $\mid \phi_{2, y} \mid + \mid \psi_{2, y} \mid = \frac{1}{3(y+2)} < 1$ for $y > 0$. Therefore this choice of rewriting the prescribed system of equations is suitable for the positive solution $(\alpha, \beta)$.

So for the root $(\tilde{\alpha}, \tilde{\beta})$ we will use the choice of $x = \phi_1(x, y)$ and $y = \psi_1(x, y)$. Using the initial approximation $(x_0, y_0) = (-1.5, -1.5)$ we have that:

(1)
\begin{align} \quad x_1 = \phi_1(x_0, y_0) = \phi_1(-1.5, -1.5) = e^{3(-1.5)} - 2 = -1.988891 \\ \quad y_1 = \psi_1(x_0, y_0) = \psi_1(-1.5, -1.5) = e^{3(-1.5)} - 2 = -1.988891 \end{align}
(2)
\begin{align} \quad x_2 = \phi_1(x_1, y_1) = \phi_1(-1.988891,-1.988891) = e^{3(-1.988891)} - 2 = -1.997437\\ \quad y_2 = \phi_1(x_1, y_1) = \psi_1(-1.988891,-1.988891) = e^{3(-1.988891)} - 2 = -1.997437 \end{align}
(3)
\begin{align} \quad x_3 = \phi_1(x_2, y_2) = \phi_1(-1.997437,-1.997437) = e^{3(-1.997437)} - 2 = -1.997502\\ \quad y_3 = \phi_1(x_2, y_2) = \psi_1(-1.997437,-1.997437) = e^{3(-1.997437)} - 2 = -1.997502 \end{align}

We have that:

(4)
\begin{align} \quad \| (x_3, y_3) - (x_2, y_2) \|_1 = \mid x_3 - x_3 \mid + \mid y_3 - y_2 \mid = 2 \mid -1.997502 - (-1.997437) \mid = 0.00013 < 0.001 = 10^{-3} = \epsilon \end{align}

Therefore $(x_3, y_3) = (-1.997502, -1.997502)$ is an approximation to the root $(\alpha, \beta)$ with accuracy $\epsilon = 10^{-3}$.