Question

The following nonlinear equations contain terms that are often found in the power flow equations:

$$\begin{gathered} {\mathbf{f}}_{\mathbf{1}}\left(x\right)\mathbf{=}\mathbf{10}{\mathbf{x}}_{\mathbf{1}}{\mathbf{sinx}}_{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{0} \\ {\mathbf{f}}_{\mathbf{2}}\left(x\right)\mathbf{=}\mathbf{10}{\left(x_1\right)}^{\mathbf{2}}\mathbf{-}\mathbf{10}{\mathbf{x}}_{\mathbf{1}}{\mathbf{cosx}}_{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0} \end{gathered}$$

Solve using the Newton-Raphson method starting with an initial guess of

${\mathbf{x}}_{\mathbf{1}}\left(0\right)\mathbf{=}\mathbf{1}$and ${\mathbf{x}}_{\mathbf{2}}\left(0\right)\mathbf{=}\mathbf{0}$ radians and a stopping criterion of $\mathbf{\epsilon }\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$.

Short answer

The solution of the system equation are$X_1=0.8553\mathrm{rad}\mathrm{and}{\mathrm{X}}_1=0.2360\mathrm{rad}$.

Step-by-step solution

Write the given data from question.

The equations of the system are,

$$\begin{gathered} {\mathrm{f}}_1\left(\mathrm{x}\right)=10{\mathrm{x}}_1{\mathrm{sinx}}_2+2=0 \\ {\mathrm{f}}_2\left(\mathrm{x}\right)=10{\left({\mathrm{x}}_1\right)}^2-10{\mathrm{x}}_1{\mathrm{cosx}}_2+1=0 \end{gathered}$$

The initial conditions,

$\begin{array}{l}{x}_1\left(0\right)=\mathrm{lrad}\\ x_2\left(0\right)=\mathrm{lrad}\end{array}$

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.

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

The equation to calculate the specific tolerance value is given as follows.

$\mathbf{\epsilon }\mathbf{=}\mathbf{\left|}\frac{\mathbf{x}\mathbf{(}\mathbf{i}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{-}{\mathbf{X}}_{\mathbf{k}}\mathbf{\left(}\mathbf{i}\mathbf{\right)}}{{\mathbf{X}}_{\mathbf{k}\mathbf{\left(}\mathbf{i}\mathbf{\right)}}}\mathbf{\right|}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{forK}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{N}$ ….. (3)

Determine the solution of the system.

Calculate the Jacobian matrix by using equation (2).

$$\begin{gathered} J{\left(i\right)}^{-1}=\left[\begin{array}{cc}\frac{a}{a{x}_1}\left(10x_1\sin x_2+2\right)& \frac{a}{a{x}_2}\left(10x_1\sin x_2+2\right)\\ \frac{a}{a{x}_1}\left(10{\left(x_1\right)}^2-10x_1\cos x_2+1\right)& \frac{a}{a{x}_2}\left(10{\left(x_1\right)}^2-10x_1\cos x_2+1\right)\end{array}\right] \\ j{\left(i\right)}^{-1}={\left[\begin{array}{cc}10\sin x_2& 10x_1\cos x_2\\ 20x_{1}10\cos x_2& 10x_1\sin x_2\end{array}\right]}^{-1} \\ j{\left(i\right)}^{-1}=\frac{\left[\begin{array}{cc}10x_1\sin x_2& 10x_1\cos x_2\\ 10\cos x_2-20x_1& 10\sin x_2\end{array}\right]}{D} \end{gathered}$$

Calculate the value of determinant of the Jacobian Matrix.

$$\begin{gathered} D=\left(10\sin x_2\right)\left(10x_1\sin x_2\right)-\left(20x_1-10\cos x_2\right)\left(10x_1\cos x_2\right) \\ D=100x_1\sin x_2-200x_1\cos x_2-100x_1\cos x_2 \\ D=100\left[\left({\sin }^2{x}_2+{\cos }^2{x}_2\right)-2{x_1}^2\cos x_2\right] \\ D=100\left[1-2{x_1}^2\cos x_2\right] \end{gathered}$$

Calculate the Jacobian matrix

localid="1655361901793" $J{\left(i\right)}^{-1}=\frac{\left[\begin{array}{c}10x_1\sin x_2-10x_1\cos x_2\\ 10\cos x_2-20x_{1}10\sin x_2\end{array}\right]}{100\left[1-2{x_1}^2\cos x_2\right]}$

Use the Newton Raphson method from equation (1).

$\left[\begin{array}{c}{x}_1\left(i+1\right)\\ x_2\left(i+1\right)\end{array}\right]=\left[\begin{array}{c}\\ \left(\begin{array}{c}{x}_1\left(i\right)\\ x_2\left(i\right)\end{array}\right)+\frac{\left[\begin{array}{c}10x_1\left(i\right)\sin x_2\left(i\right)-10x_1\left(i\right)\cos x_2\left(i\right)\\ 10\cos x_2\left(i\right)-20x_1\left(i\right)10\sin x_2\left(i\right)\end{array}\right]}{100\left[1-2{x_1}^2\left(i\right)\cos x_2\right]}\times \\ \left[\begin{array}{c}-10x_1\left(i\right)10\sin x_2\left(i\right)-2\\ -10{\left(x_1\right)}^2\left(i\right)+10x_1\left(i\right)\cos x_2\left(i\right)-1\end{array}\right]\\ \end{array}\right]$

The above equation can be written as two separate equation,

$x_1\left(i+1\right)x_1\left(i\right)+\frac{\left\{\begin{array}{c}\left(10x_1\left(i\right)\sin x_2\left(i\right)\right)\left(10x_1\left(i\right)\sin x_2\left(i\right)-2\right)\\ +\left(-10x_1\left(i\right)\cos x_2\left(i\right)\right)\left(-10{\left(x_1\right)}^2\left(i\right)\cos x_2\left(i\right)-1\right)\end{array}\right\}}{100\left[1-2{x_1}^2\left(i\cos x_2\right)\right]}$ …… (4)

localid="1655360578288" $x_2\left(i+1\right)=x_2\left(i\right)+\frac{\left\{\begin{array}{c}\left(10\cos x_2\left(i\right)-20x_1\left(i\right)\right)\left(-10x_1\left(i\right)\sin x_2\left(i\right)-2\right)\\ +\left(10\sin x_2\left(i\right)\right)\left(-10{\left(x_1\right)}^2\left(i\right)+10x_1\left(i\right)\cos x_2\left(i-1\right)\right)\end{array}\right\}}{100\left[1-2{x_1}^2\left(i\right)\cos x_2\right]}$ …… (5)

Calculate the specific tolerance value.

localid="1655360584035" $\mathrm{\epsilon }=\left|\frac{{\mathrm{X}}_{\mathrm{k}}(\mathrm{i}+1)-{\mathrm{X}}_{\mathrm{k}}\left(\mathrm{i}\right)}{{\mathrm{X}}_{\mathrm{k}}\left(\mathrm{i}\right)}\right|\mathrm{forK}=1,2...\mathrm{N}$ ……. (6)

The table for the successive iteration calculation of the equation (4), (5) and (6).

Newton Raphson method converges after 4 iteration and repeating the values. The specific tolerance value is less than ${10}^{-4}$.

Hence the solution of the system equation are${\mathrm{X}}_1=0.8553\mathrm{rad}\mathrm{and}\mathrm{x}=-0.2360\mathrm{rad}$.