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
\[
J=
\left(
\begin{array}{cc}
J^1_\varphi & J^1_A \ J^2_\varphi & J^2_A
\end{array}
\right)
\]
Levenberger-Marquardt for nonlinear elliptical system
原文：https://www.cnblogs.com/yuewen-chen/p/11525996.html