Question

Using the Jacobi method (also known as the Gauss method), solve

for${\mathbf{x}}_{\mathbf{1}}$and${\mathbf{x}}_{\mathbf{2}}$in the following system of equations.

$$\begin{gathered} {\mathbf{x}}_{\mathbf{2}}\mathbf{-}\mathbf{3}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{9}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}{\mathbf{x}}_{\mathbf{2}}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{=}\mathbf{0} \end{gathered}$$

Use an initial guess of ${\mathbf{x}}_{\mathbf{1}}\left(0\right)\mathbf{=}\mathbf{1}$and ${\mathbf{x}}_{\mathbf{2}}\left(0\right)\mathbf{=}\mathbf{1}$. Also, see what happens when

you choose an uneducated initial guess of${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{100}$.

Short answer

The solution of the system with initial condition ${\mathrm{x}}_1\left(0\right)={\mathrm{x}}_2\left(0\right)=1$are $x_1=0.94$and

$x_2=0.92$. The solution of system does not converge with uneducated initial guess.

Step-by-step solution

Write the given data from question.

The system equations are,

$$\begin{gathered} x_2-3x_1+1.9=0 \\ {x}_2+x_1^2-1.8=0 \end{gathered}$$

The initial condition,

$$\begin{gathered} x_1\left(0\right)=1 \\ {x}_2\left(0\right)=1 \end{gathered}$$

The uneducated initial guess,

${\mathrm{x}}_1\left(0\right)={\mathrm{x}}_2\left(0\right)=100$

Determine the solution of the system and effect of uneducated initial condition on system.

Rewrite the system equations as,

$$\begin{gathered} x_1=\frac{1}{3}{x}_2+0.63 \\ {x}_2=-x_1^2+1.8 \end{gathered}$$

The system equation for the Jacobi method,

$$\begin{gathered} x_1\left(i+1\right)=\frac{1}{3}{x}_2\left(i\right)+0.63.......\left(1\right) \\ {x}_2\left(i+1\right)=1.8-x_2^1\left(i+1\right).......\left(2\right) \end{gathered}$$

Calculate the first iteration for $x_1$and $x_2$.

Substitute 1 for $x_1\left(0\right)$into equation (1).

$$\begin{gathered} x_1\left(1\right)=\frac{1}{3}\left(1\right)+0.63 \\ {x}_1\left(1\right)=0.96 \end{gathered}$$

Substitute 0.96 for$x_1\left(1\right)$into equation (2).

$$\begin{gathered} x_2\left(1\right)=1.8-0.{96}^2 \\ {x}_2\left(1\right)=0.8784 \end{gathered}$$

Determine the second iteration$x_1$and$x_2$.

For$x_1$

$x_1\left(2\right)=\frac{1}{3}{x}_2\left(1\right)+0.63$

Substitute 0.8784 for$x_2\left(1\right)$into above equation.

$$\begin{gathered} x_1\left(2\right)=\frac{1}{3}\left(0.8784\right)+0.63 \\ {x}_1\left(2\right)=0.92 \end{gathered}$$

For$x_2$

$x_2\left(2\right)=1.8-x_1^2\left(2\right)$

Substitute 0.92 for$x_2\left(1\right)$into above equation.

$$\begin{gathered} x_1\left(2\right)=\frac{1}{3}\left(0.8784\right)+0.63 \\ {x}_1\left(2\right)=0.92 \end{gathered}$$

Determine the third iteration$x_1$and$x_2$.

For$x_1$

$x_1\left(2\right)=\frac{1}{3}{x}_2\left(2\right)+0.63$

Substitute 0.954 for$x_2\left(2\right)$into above equation.

$$\begin{gathered} x_1\left(3\right)=\frac{1}{3}\left(0.954\right)+0.63 \\ {x}_1\left(3\right)=0.94 \end{gathered}$$

For$x_2$

$x_2\left(3\right)=1.8-x_1^2\left(3\right)$

Substitute 0.94 for$x_1\left(3\right)$into above equation.

$$\begin{gathered} x_2\left(3\right)=1.8-x_1^2 \\ {x}_2\left(3\right)=0.92 \end{gathered}$$

Determine the fourth iteration$x_1$and$x_2$.

Forrole="math" localid="1655370670224" $x_1$

$x_1\left(4\right)=\frac{1}{3}{x}_2\left(3\right)+0.63$

Substitute 0.94 for $x_2\left(3\right)$into above equation.

$$\begin{gathered} x_1\left(4\right)=\frac{1}{3}\left(0.94\right)+0.63 \\ {x}_1\left(4\right)=0.94 \end{gathered}$$

For$x_2$

Substitute 0.94 for$x_1\left(4\right)$into above equation.

$$\begin{gathered} x_2\left(4\right)=1.8-0.{94}^2 \\ {x}_2\left(4\right)=0.92 \end{gathered}$$

Since the iteration converges after the fourth iteration. Therefore the solution of the

system’s equation are$x_1=0.94and{x}_2-0.92$

Determine the first iteration for uneducated initial condition.

Determine the first iteration $x_1$and $x_2$.

For $x_1$

$x_1\left(1\right)=\frac{1}{3}{x}_2\left(0\right)+0.63$

Substitute 100 for $x_2\left(0\right)$into above equation.

$$\begin{gathered} x_1\left(2\right)=\frac{1}{3}{x}_2\left(100\right)+0.63 \\ {x}_1\left(2\right)=33.96 \end{gathered}$$

For$x_2$

$x_2\left(1\right)=1.8-x_1^2\left(1\right)$

Substitute 33.96 for $x_1\left(1\right)$into above equation.

$$\begin{gathered} x_2\left(2\right)=1.8-33.{96}^2 \\ {x}_2\left(2\right)-1151 \end{gathered}$$

Determine the second iteration$x_1$and$x_2$.

Forrole="math" localid="1655372518984" $x_1$

$x_1\left(2\right)=\frac{1}{3}{x}_2\left(1\right)+0.63$

Substitute -1151.48 for$x_2\left(1\right)$into above equation.

$$\begin{gathered} x_1\left(2\right)=\frac{1}{3}\left(-1151.48\right)+0.63 \\ {x}_1\left(2\right)=-383.19 \end{gathered}$$

For$x_2$

Substitute -383.19 for$x_1\left(2\right)$into above equation.

$$\begin{gathered} x_2\left(2\right)=1.8{\left(-383.19\right)}^2 \\ {x}_2\left(2\right)=-146832.77 \end{gathered}$$

From the above the values are increased by vary large number. Therefore, the

solution of system does not converge with uneducated initial guess.

Hence, the solution of the system with initial condition $x_1\left(0\right)=x_2\left(0\right)=1$are $x_1=0.94$and $x_2=0.92$. The solution of system does not converge with uneducated

initial guess.