Question 3.5: Solution of a system of nonlinear equations using Newton's m......

Equations (3.49) and (3.50) are a system of two nonlinear equations. The points of intersection are given by the solution of the system. The solution with Newton’s method is obtained by using Eqs. (3.43) and (3.44). In the present problem, the partial derivatives in the equations are given by:

\frac{\partial f_{1}}{\partial x}|_{x_1,y_1} \Delta x + \frac{\partial f_{1}}{\partial y}|_{x_1,y_1} \Delta y = – f_1(x_1,y_1)        (3.43)

\frac{\partial f_{2}}{\partial x}|_{x_1,y_1} \Delta x + \frac{\partial f_{2}}{\partial y}|_{x_1,y_1} \Delta y = – f_2(x_1,y_1)        (3.44)

\frac{\partial f_{1}}{\partial x}=-\frac{1}{4}\left(e^{x / 2}-e^{(-x) / 2}\right) \quad \text { and } \quad \frac{\partial f_{1}}{\partial y}=1        (3.51)

\frac{\partial f_{2}}{\partial x}=18 x \text { and } \frac{\partial f_{2}}{\partial y}=50 y       (3.52)

The Jacobian is given by:

\left.J\left(f_{1}, f_{2}\right)=\operatorname{det} \left[\begin{array}{ll}\frac{\partial f_{1}}{\partial x} & \frac{\partial f_{1}}{\partial y} \\\frac{\partial f_{2}}{\partial x} & \frac{\partial f_{2}}{\partial y}\end{array}\right]\right.=\operatorname{det}\left[\begin{array}{cc} -\frac{1}{4}\left(e^{x / 2}-e^{(-x) / 2}\right) & 1 \\ 18 x & 50 y \end{array}\right]=-\frac{1}{4}\left(e^{x / 2}-e^{(-x) / 2}\right) 50 y-18 x         (3.53)

Substituting Eqs. (3.51)-(3.53) in Eqs. (3.45) and (3.46) gives the solution for \Delta x and \Delta y.

\Delta x = \frac{ – f_1(x_1,y_1) \frac{\partial f_{2}}{\partial y}|_{x_1,y_1} + f_2(x_1,y_1) \frac{\partial f_{1}}{\partial y}|_{x_1,y_1}}{J(f_1(x_1,y_1),f_2(x_1,y_1))}       (3.45)

\Delta y = \frac{ – f_2(x_1,y_1) \frac{\partial f_{1} (x_1,y_1)}{\partial x}|_{x_1,y_1} + f_1(x_1,y_1) \frac{\partial f_{2}}{\partial x}|_{x_1,y}}{J(f_1(x_1,y_1),f_2(x_1,y_1))}       (3.46)

The problem is solved in the MATLAB program that is listed below. The order of operations in the program is:

• The solution is initiated by the initial guess, x_{i}=2.5, y_{i}=2.0.
• The iterations start. \Delta y and \Delta x are determined by substituting x_{i} and y_{i} in Eqs. (3.45) and (3.46).
• x_{i+1}=x_{i}+\Delta x, and y_{i+1}=y_{i}+\Delta y are determined.
• If the estimated relative error (Eq. (3.9)) for both variables is smaller than 0.001 , the iterations stop. Otherwise, the values of x_{i+1} and y_{i+1} are assigned to x_{i} and y_{i}, respectively, and the next iteration starts.

EstimatedRelativeError = \left| \frac{x_{NS}^{(n)}  –  x_{NS}^{(n  –  1)}}{x_{NS}^{(n  –  1)}}\right|       (3.9)

The program also displays the solution and the error at each iteration.