Question

For the Newton-Raphson method, the region of attraction (or basin of attraction) for a particular solution is the set of all initial guesses that converge to that solution. Usually initial guesses close to a particular solution will converge to that solution. However, for all but the simplest of multi-dimensional, nonlinear problems, the region of attraction boundary is often fractal. This makes it impossible to quantify the region of attraction and hence to guarantee convergence. Problem 6.25 has two solutions when x2 is restricted to being between$-\pi$and$\pi$. With the$x_2$initial guess fixed at$0$radians, numerically determine the values of theinitial guesses that converge to the Problem 6.25 solution. Restrict your search to values $x_1$of $0$betweenand$1$.

Short answer

The value of the$x_1$ is $0.8553$that lies between$0$and$1$at the 5^th iteration.

Step-by-step solution

Write the given data from question.

The equations of the system are,

$\begin{array}{l}{f}_1\left(x\right)=10x_1\sin x_2+2=0\\ f_2\left(x\right)=10{\left(x_1\right)}^2-10x_1\cos x_2+1=0\end{array}$

The initial conditions,

$\begin{array}{l}{x}_1\left(0\right)=1\text{\hspace{0.17em}rad}\\ x_2\left(0\right)=0\text{\hspace{0.17em}rad}\end{array}$

Determine the equation to calculate the value of between 0 and 1.

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

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

The Jacobian matrix is given as follows.

$J{\left(i\right)}^{-1}={\left[\begin{array}{cc}\frac{\partial f_1}{\partial x_1}& \frac{\partial f_1}{\partial x_2}\\ \frac{\partial f_2}{\partial x_1}& \frac{\partial f_2}{\partial x_2}\end{array}\right]}^{-1}$ …… (2)

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

$\epsilon =\left|\frac{x_k\left(i+1\right)-x_k\left(i\right)}{x_k\left(i\right)}\right|forK=1,2,...N$ ….. (3)

Determine the iteration in which x1 lies between 0 and 1.

Calculate the Jacobian matrix by using equation (2).

$\begin{array}{l}J{\left(i\right)}^{-1}={\left[\begin{array}{cc}\frac{\partial }{\partial x_1}\left(10x_1\sin x_2+2\right)& \frac{\partial }{\partial x_2}\left(10x_1\sin x_2+2\right)\\ \frac{\partial }{\partial x_1}\left(10{\left(x_1\right)}^2-10x_1\cos x_2+1\right)& \frac{\partial }{\partial x_2}\left(10{\left(x_1\right)}^2-10x_1\cos x_2+1\right)\end{array}\right]}^{-1}\\ J{\left(i\right)}^{-1}={\left[\begin{array}{cc}10\sin x_2& 10x_1\cos x_2\\ 20x_1-10\cos x_2& 10x_1\sin x_2\end{array}\right]}^{-1}\\ J{\left(i\right)}^{-1}=\frac{\left[\begin{array}{cc}10x_1\sin x_2& -10x_1\cos x_2\\ 10\cos x_2-20x_1& 10\sin x_2\end{array}\right]}{D}\end{array}$

Here$D$is determinant of the Jacobian matrix.

Calculate the value of determinant of theJacobian Matrix.

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

Calculate the Jacobian matrix

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

Use the Newton Raphson method from equation (1).

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

The above equation can be written as two separate equation,

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

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

Calculate the specific tolerance value.

$\epsilon =\left|\frac{x_k\left(i+1\right)-x_k\left(i\right)}{x_k\left(i\right)}\right|\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}for\hspace{0.17em}K\hspace{0.17em}=\hspace{0.17em}1,2,}\mathrm{....}\text{N}$ ……. (6)

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

$i$	$x_1\left(i\right)$	$x_2\left(i\right)$	$\epsilon$
0	1	0	0.1
1	0.9	-0.2	0.1666
2	0.8586	-0.2333	0.0113
3	0.8553	-0.2360
4	0.8553	-0.2360

From the above solution the value of the$x_1$ lies between$0$and$1$ at the 5^th.

Hence, the value of the$x_1$is$0.8553$that lies between$0$and$1$at the 5^th iteration.