Problem:
Find the zeros of the following system of equations using Newton's method. \begin{eqnarray*} f(x,y) &=& x^2 + y^2 - 3 \\ g(x,y) &=& 3x^2 + y^3 - 6 \\ \end{eqnarray*}
Answer:
I am using the following web page as a source for Newton's method.
www.math.ohiou.edu/courses/math3600/lecture13.pdf
The following plot shows that there is at least one solution. [Graph][http://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+y%5E2+-+3+%3D+0+,+3*x%5E2+%2B+y%5E3+-+6+%3D+0]
Note: This Plot only shows one equation which is less than ideal.
First, I find the partial derivatives. \begin{eqnarray*} f_x &=& 2x \\ f_y &=& 2y \\ g_x &=& 6x \\ g_y &=& 3y^2 \\ \end{eqnarray*} My initial guess with be $x_0 = [1,1]$. Note that $x_0$ is a vector with two components $(x,y)$. Now, I apply the formula: \begin{eqnarray*} x_1 &=& x_0 - (Df(x_0))^{-1} f(x_0) \\ Df(x_0) &=& \begin{bmatrix} f_x(1,1) & f_y(1,1) \\ g_x(1,1) & g_y(1,1) \\ \end{bmatrix} \\ Df(x_0) &=& \begin{bmatrix} 2 & 2 \\ 6 & 3 \\ \end{bmatrix} \\ \end{eqnarray*} Now, I have a problem. This matrix is not invertible. Did I do something wrong? Is there a reason why this method would not work correctly in this case? Bob
Correction: After fixing my error, I get an invertible matrix. Thank you.
You have a mistake in your partial derivatives as $g_x = 6x$ in which case the Jacobian at $x_0 = [1,1]$ is not singular and converges to $x = [1.0885, 1.3473]$. Otherwise, singular Jacobians typically arise from a bad initial starting point. There is a modified Newton's method which can handle this, see Hueso, J. L., et al., Modified Newton's method for systems of nonlinear equations with singular Jacobian.