Question

Solve the following equations by the Newton-Raphson method:

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

Start with an initial guess of ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{1}$

Short answer

The solution of the equation of the system are$x_1=1.8560\mathrm{and}{x}_2=2.0707$.

Step-by-step solution

Write the given data from question.

The equations of the system are

$$\begin{gathered} f_1=2{\mathrm{x}}_1+{{\mathrm{x}}_2}^2-8=0 \\ {{\mathrm{f}}_2={\mathrm{x}}_1}^2-{{\mathrm{x}}_2}^2+{\mathrm{x}}_1{\mathrm{x}}_2-3=0 \end{gathered}$$

The initial conditions,

$\begin{array}{l}{X}_1\left(0\right)=1\\ X_2\left(0\right)=1\end{array}$

Determine the equation to calculate the solution of the equation.

The generalise form of the Newton-Raphson method is given as follows.

$\mathbf{x}\mathbf{(}\mathbf{i}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{x}\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{+}\mathbf{J}{\mathbf{\left(}\mathbf{i}\mathbf{\right)}}^{\mathbf{-}\mathbf{1}}\mathbf{[}\mathbf{y}\mathbf{-}\mathbf{f}\mathbf{\left(}\mathbf{x}\left(i\right)\mathbf{\right)}\mathbf{]}$ …… (1)

The Jacobian matrix is given as follows.

$\mathbf{J}{\mathbf{\left(}\mathbf{i}\mathbf{\right)}}^{\mathbf{-}\mathbf{1}}\mathbf{=}{\mathbf{\left[}\begin{array}{cc}\frac{{\mathbf{af}}_{\mathbf{1}}}{{\mathbf{ax}}_{\mathbf{1}}}& \frac{{\mathbf{af}}_{\mathbf{1}}}{{\mathbf{ax}}_{\mathbf{2}}}\\ \frac{{\mathbf{af}}_{\mathbf{2}}}{{\mathbf{ax}}_{\mathbf{1}}}& \frac{{\mathbf{af}}_{\mathbf{2}}}{{\mathbf{ax}}_{\mathbf{2}}}\end{array}\mathbf{\right]}}^{\mathbf{-}\mathbf{1}}$

Determine the solution of the system.

Calculate the Jacobian matrix by using equation (2).

$$\begin{gathered} J{\left(i\right)}^{-1}=\left[\begin{array}{cc}\frac{a}{a{x}_1}\left(2x_1+{x_2}^2-8\right)& \frac{a}{a{x}_2}\left(2x_1+{x_2}^2-8\right)\\ \frac{a}{a{x}_1}\left({x_1}^2-{x_2}^2+x_1{x}_2-3\right)& \frac{a}{a{x}_2}\left({x_1}^2-{x_2}^2+x_1{x}_2-3\right)\end{array}\right] \\ J{\left(i\right)}^{-1}=\left[\begin{array}{cc}2& 2x_2\\ 2x_1+x_2& -2x_2+x_1\end{array}\right] \\ J{\left(i\right)}^{-1}=\frac{\left[\begin{array}{cc}-2x_2+x_1& -2x_2\\ -2x_1-x_2& 2\end{array}\right]}{D} \end{gathered}$$

Here $D$is determinant of the Jacobian matrix.

Calculate the value ofdeterminant of the Jacobian Matrix.

$\begin{array}{l}D=\left(2\right)\left(-2_1+X_2\right)-\left(2X_2\right)(2X_1+X_2^{})\\ D=-4X_X+2X_2-4X_2{x}_1-2{X^2}_2\end{array}$

Calculate the Jacobian matrix

$J{\left(i\right)}^{-1}=\frac{\left[\begin{array}{cc}-2X_2+X_1& -2X_2\\ -X_1-X_2& 2\end{array}\right]}{4X_{+}+2x_2-4X_1{X}_2-2X_2^2}$

Use the Newton Raphson method from equation (1).

localid="1655353968588" $\left[\begin{array}{c}{x}_1\left(i+1\right)\\ x_2\left(i+1\right)\end{array}\right]=\left[\begin{array}{c}\left(\begin{array}{c}\begin{array}{c}{x}_1\left(i\right)\\ x_2\left(i\right)\end{array}\end{array}\right)+\frac{\left[\begin{array}{cc}-2x_2\left(i\right)+x_1\left(i\right)& -2x_2\left(i\right)\\ -2x_1\left(i\right)-x_2\left(i\right)& 2\end{array}\right]}{-4x_1\left(i\right)+2x_2\left(i\right)+4x_1{x}_2\left(i\right)+2{x^2}_2\left(i\right)}\\ \times \left[\begin{array}{c}8-2x_1\left(i\right)-{x^2}_2\left(i\right)\\ 3-{x_1}^2\left(i\right)+2{x^2}_2\left(i\right)-x_1{x}_2\left(i\right)\end{array}\right]\end{array}\right]$

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

localid="1655355465620" $\left[\begin{array}{c}{x}_1\left(0+1\right)\\ x_2\left(0+1\right)\end{array}\right]=\left[\begin{array}{c}\left(\begin{array}{c}\begin{array}{c}{x}_1\left(0\right)\\ x_2\left(0\right)\end{array}\end{array}\right)+\frac{\left[\begin{array}{cc}-2x_2\left(0\right)+x_1\left(0\right)& -2x_2\left(0\right)\\ -2x_1\left(0\right)-x_2\left(0\right)& 2\end{array}\right]}{-4x_1\left(0\right)+2x_{20}\left(0\right)+4x_1{x}_2\left(0\right)+2{x^2}_2\left(0\right)}\\ \times \left[\begin{array}{c}8-2x_1\left(0\right)-{x^2}_2\left(0\right)\\ 3-{x_1}^2\left(0\right)+{x^2}_2\left(0\right)-x_1{x}_2\left(0\right)\end{array}\right]\end{array}\right]$

localid="1655355471183" $\left[\begin{array}{c}{x}_1\left(1\right)\\ x_2\left(1\right)\end{array}\right]=\left[\begin{array}{c}\left(\begin{array}{c}\begin{array}{c}{x}_1\left(0\right)\\ x_2\left(0\right)\end{array}\end{array}\right)+\frac{\left[\begin{array}{cc}-2x_2\left(0\right)+x_1\left(0\right)& -2x_2\left(0\right)\\ -2x_1\left(0\right)-x_2\left(0\right)& 2\end{array}\right]}{-4x_1\left(0\right)+2x_{20}\left(0\right)+4x_1{x}_2\left(0\right)+2{x^2}_2\left(0\right)}\\ \times \left[\begin{array}{c}8-2x_1\left(0\right)-{x^2}_2\left(0\right)\\ 3-{x_1}^2\left(0\right)+{x^2}_2\left(0\right)-x_1{x}_2\left(0\right)\end{array}\right]\end{array}\right]$

Substitute localid="1655355483574" $1$for localid="1655355476471" $x_1\left(0\right)\mathrm{and}1for{x}_2\left(0\right)$into above equation.

localid="1655355548432" $$\begin{gathered} \left[\begin{array}{c}{x}_1\left(1\right)\\ x_2\left(1\right)\end{array}\right]=\left[\begin{array}{c}\left(\begin{array}{c}\begin{array}{c}1\\ 1\end{array}\end{array}\right)+\frac{\left[\begin{array}{cc}-2x_2\left(1\right)+x_1\left(1\right)& -2x_2\left(1\right)\\ -2x_1\left(1\right)-x_2\left(1\right)& 2\end{array}\right]}{-4x_1\left(1\right)+2x_{20}\left(1\right)+4x_1{x}_2\left(1\right)+2{x^2}_2\left(1\right)}\\ \end{array}\times \left[\begin{array}{c}8-2x_1\left(1\right)-{x^2}_2\left(1\right)\\ 3-{x_1}^2\left(1\right)+{x^2}_2\left(0\right)-x_1{x}_2\left(1\right)\end{array}\right]\right] \\ \left[\begin{array}{c}{x}_1\left(1\right)\\ x_2\left(1\right)\end{array}\right]=\left(\begin{array}{c}\begin{array}{c}1\\ 1\end{array}\end{array}\right)+\frac{\left[\begin{array}{cc}-1& -2\\ -3& 2\end{array}\right]}{\left(-8\right)}\times \left[\begin{array}{c}5\\ 2\end{array}\right] \\ \left[\begin{array}{c}{x}_1\left(1\right)\\ x_2\left(1\right)\end{array}\right]=\left(\begin{array}{c}\begin{array}{c}1\\ 1\end{array}\end{array}\right)+\frac{\left[\begin{array}{cc}-5& -4\\ -15& 4\end{array}\right]}{-8} \\ \left[\begin{array}{c}{x}_1\left(1\right)\\ x_2\left(1\right)\end{array}\right]=\left(\begin{array}{c}\begin{array}{c}1\\ 1\end{array}\end{array}\right)+\frac{\left[\begin{array}{c}-9\\ -11\end{array}\right]}{-8} \end{gathered}$$

Solve further as,

localid="1655355563144" $$\begin{gathered} \left[\begin{array}{c}{x}_1\left(1\right)\\ x_2\left(1\right)\end{array}\right]=\left(\begin{array}{c}\begin{array}{c}1\\ 1\end{array}\end{array}\right)+\frac{\left[\begin{array}{c}9\\ 11\end{array}\right]}{8} \\ \left[\begin{array}{c}{x}_1\left(1\right)\\ x_2\left(1\right)\end{array}\right]=\left(\begin{array}{c}\begin{array}{c}1\\ 1\end{array}\end{array}\right)+\left[\begin{array}{c}1.125\\ 1.375\end{array}\right] \\ \left[\begin{array}{c}{x}_1\left(1\right)\\ x_2\left(1\right)\end{array}\right]=\left[\begin{array}{c}2.125\\ 2.375\end{array}\right] \end{gathered}$$

The table for the successive iteration is shown below.

The value of $x_1\left(4\right)$is $1.8560$and $x_2\left(4\right)$is $2.0707$.

Since the value of$X_1\mathrm{and}{\mathrm{X}}_2$ are repeating. Therefore, the solution of the equation of the system are$X_1=1.8560\mathrm{and}{\mathrm{x}}_2=2.0707$.