Question

Use Newton-Raphson to find a solution to

$\left[\begin{array}{c}{e}^{x_1{x}_2}\\ \cos \left(x_1+x_2\right)\end{array}\right]\mathbf{=}\left[\begin{array}{c}1.2\\ 0.5\end{array}\right]$

where${\mathbf{x}}_{\mathbf{1}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{x}}_{\mathbf{2}}$are in radians. (a) Start with${\mathbf{x}}_{\mathbf{1}}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{x}}_{\mathbf{2}}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$and continue until (6.2.2) is satisfied with$\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{005}$. (b) Show that Newton-Raphson diverges for this example if${\mathbf{x}}_{\mathbf{1}}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{x}}_{\mathbf{2}}\left(0\right)\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}$.

Short answer

(a) The solution of the system are ${\mathrm{x}}_1=0.8266\mathrm{and}{\mathrm{x}}_2=0.2205$

(b) It is shown that the Newton Raphson method diverges if initial conditions are${\mathrm{x}}_1\left(0\right)=1\mathrm{and}{\mathrm{x}}_2\left(0\right)=2.0.$

Step-by-step solution

Write the given data from question.

Consider the given matrix.

$\left[\begin{array}{c}{\mathrm{e}}^{{\mathrm{x}}_1{\mathrm{x}}_2}\\ \cos ({\mathrm{x}}_1+{\mathrm{x}}_2)\end{array}\right]=\left[\begin{array}{c}1.2\\ 0.5\end{array}\right]$

Here,

$\begin{array}{l}{f}_1\left(x\right)=e^8+\\ f_2\left(x\right)=\cos \left(x_1+X_2\right)\end{array}$

Consider the initial conditions.

$\begin{array}{l}{X}_1\left(0\right)=1\\ X_2\left(0\right)=0.5\end{array}$

Determine the equation to calculate the solution of the equation.

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

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

The Jacobian matrix is given as follows.

$\mathbf{J}{\left(i\right)}^{\mathbf{-}\mathbf{1}}\mathbf{=}\left[\begin{array}{cc}\frac{{\mathrm{af}}_1}{{\mathrm{ax}}_1}& \frac{{\mathrm{af}}_1}{{\mathrm{ax}}_2}\\ \frac{{\mathrm{af}}_2}{{\mathrm{ax}}_1}& \frac{{\mathrm{af}}_2}{{\mathrm{ax}}_2}\end{array}\right]$ …… (2)

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

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

Determine the solution of the system.

(a)

Calculate the Jacobian matrix by using equation (2).

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

Here, $D$is determinant of the Jacobian matrix.

Calculate the value of determinant of theJacobian Matrix.

$$\begin{gathered} D=\left(\sin \left[x_1\left(i\right)+x_2\left(i\right)\right]\right)\left(x_2\left(i\right)e^{x_1{x}_2\left(i\right)}\right)-\left(\sin \left[x_1\left(i\right)+x_2\left(i\right)\right]\right)\left(-x_1\left(i\right)e^{x_1\left(i\right)x_2\left(i\right)}\right) \\ D=\sin \left[x_1\left(i\right)+x_2\left(i\right)\right]e^{x_1\left(i\right)x_2\left(i\right)}\left[x_1\left(i\right)-x_2\left(i\right)\right] \end{gathered}$$

Calculate the Jacobian matrix.

$J{\left(i\right)}^{-1}=\frac{\left[\begin{array}{c}-\sin \left[x_1\left(i\right)+x_2\left(i\right)\right]\left(-x_1\left(i\right)e^{x_1\left(i\right)x_2\left(i\right)}\right)\\ \sin \left[x_1\left(i\right)+x_2\left(i\right)\right]\left(-x_1\left(i\right)e^{x_1\left(i\right)x_2\left(i\right)}\right)\end{array}\right]}{\sin \left[x_1\left(i\right)+x_2\left(i\right)\right]e^{x_1\left(i\right)x_2\left(i\right)}{x}_1\left(i\right)-x_2\left(i\right)}$

Use the Newton Raphson method from equation (1).

$\left[\begin{array}{c}{x}_1\left(i+1\right)\\ x_2\left(i+1\right)\end{array}\right]=\left[\begin{array}{c}\\ \left(\begin{array}{c}{x}_1\left(i\right)\\ x_2\left(i\right)\end{array}\right)+\frac{\left[\begin{array}{c}\sin \left[x_1\left(i\right)+x_2\left(i\right)\right]x_1\left(i\right)-x_2\left(i\right)\\ \sin \left[x_1\left(i\right)+x_2\left(i\right)\right]x_1\left(i\right)-x_2\left(i\right)\end{array}\right]}{\sin \left[x_1\left(i\right)+x_2\left(i\right)\right]x_1\left(i\right)-x_2\left(i\right)}\\ \times \left[\begin{array}{c}1.2-e^{x_1\left(i\right)x_2\left(i\right)}\\ 0.5-\left(\cos \left(x_1\left(i\right)+x_2\left(i\right)\right)\right)\end{array}\right]\\ \end{array}\right]$

The above equation can be written as two separate equation,

$x_1\left(i+1\right)=\left[x_1\left(i\right)+\frac{\begin{array}{c}-\sin \left[x_1\left(i\right)+x_2\left(i\right)\right]\left(1.2-e^{x_1\left(i\right)x_2\left(i\right)}\right)\\ -x_1\left(i\right)e^{x_1\left(i\right)x_2\left(i\right)}\left(0.5-\left(\cos \left(x_1\left(i\right)+x_2\left(i\right)\right)\right)\right)\end{array}}{\sin \left[x_1\left(i\right)+x_2\left(i\right)\right]e^{x_1\left(i\right)x_2\left(i\right)}\left[x_1\left(i\right)+x_2\left(i\right)\right]}\right]$ …… (4)

$x_2\left(i+1\right)=\left[x_2\left(i\right)+\frac{\begin{array}{c}\sin \left[x_1\left(i\right)+x_2\left(i\right)\right]\left(1.2-e^{x_1\left(i\right)x_2\left(i\right)}\right)\\ x_2\left(i\right)e^{x_1\left(i\right)x_2\left(i\right)}\left(0.5-\left(\cos \left(x_1\left(i\right)+x_2\left(i\right)\right)\right)\right)\end{array}}{\sin \left[x_1\left(i\right)+x_2\left(i\right)\right]e^{x_1\left(i\right)x_2\left(i\right)}\left[x_1\left(i\right)+x_2\left(i\right)\right]}\right]$ …… (5)

Calculate the specific tolerance value.

$\epsilon =\left|\frac{x_k\left(i+1\right)-x_k\left(i\right)}{x_k\left(i\right)}\right|\mathrm{forK}=1,2....\mathrm{N}$

The table of the successive iteration of the above three equation.

The value of specific tolerance is converging in four iterations and less than $0.005$.

Hence the solution of the system are$x_1=0.8266\mathrm{and}{x}_2=0.2205$

Determine that Newton-Raphson diverges for this example if x1(0)=1.0 and x2(0)=2.0.

(b)

The initial condition for given system equation are$x_1\left(0\right)=1\mathrm{and}{\mathrm{x}}_2\left(0\right)=2.0$.

The table of the successive iteration of the equations (4), (5) and (6).

From the above table the value of $\epsilon$is increasing rapidly and diverges.

Hence, it is shown that the Newton Raphson method diverges if initial conditions are$x_1\left(0\right)=1\mathrm{and}{\mathrm{x}}_2\left(0\right)=2.0$.