Levenberger-Marquardt for nonlinear elliptical system

考虑如下的方程组，测试Levenberger-Marquardt 方法：\
\begin{align}
\varphi_{rr}+\frac{2}{r}\varphi_{,r}+\frac{1}{8}(A)^2-\frac{1}{12}\varphi^5 &=f_1\
A_{,r}-\frac{2}{3}\varphi^6+A &=f_2\
f_1&=\frac{r^2}{2}-\frac{2}{r (r+1)^2}+\frac{2}{(r+1)^3}-\frac{1}{12 (r+1)^5}\
f_2&=2 r-\frac{2}{3 (r+1)^6}+2
\end{align}
使用中心差分，以下计算出他们的Jacobi 矩阵：\
\begin{align}
a_{i+1}&= F^1_{\varphi_{i+1}} =\frac{1}{h^2}+\frac{1}{h r(i)}\
a_i &= F^1_{\varphi_i} =-\frac{2}{h^2}-\frac{1}{12} 5 \varphi(i)^4 \
a_{i-1}&= F^1_{\varphi_{i-1}} = \frac{1}{h^2}-\frac{1}{h r(i)}\
b_{i} &= F^1_{A_{i}} =\frac{A(i)}{4} \
c_{i} &= F^2_{\varphi_{i}} = -4 \varphi(i)^5 \
d_{i+1}&= F^2_{A_{i+1}} =\frac{1}{2h} \
d_{i} &= F^2_{A_{i}} =1 \
d_{i-1}&= F^2_{A_{i-1}} =-\frac{1}{2h} \
\end{align}
\[ J^1_\varphi= \left( \begin{array}{ccc} a_1 & a_2 & 0 \\ a_1 & a_2 & a_3 \\ 0 & a_{n-1} & a_n \\ \end{array} \right) \]

\[ J^1_A= \left( \begin{array}{ccc} b_1 & 0 & 0 \\ 0 & b_2 & 0 \\ 0 & 0 & b_n \\ \end{array} \right) \]

\[ J^2_\varphi= \left( \begin{array}{ccc} c_1 & 0 & 0 \\ 0 & c_2 & 0 \\ 0 & 0 & c_n \\ \end{array} \right) \]

\[ J^2_A= \left( \begin{array}{ccc} d_1 & d_2 & 0 \\ d_1 & d_2 & d_3 \\ 0 & d_{n-1} & d_n \\ \end{array} \right) \]
完整的jacobi 是以上四个构成的分块矩阵
\[ J= \left( \begin{array}{cc} J^1_\varphi & J^1_A \\ J^2_\varphi & J^2_A \end{array} \right) \]

来源：https://www.cnblogs.com/yuewen-chen/p/11525996.html