解常微分方程组的问题 - 未名空间MITBBS历史存档

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解常微分方程组的问题

解常微分方程组的问题# Computation - 科学计算

l*y2007-06-08 07:06

1 楼

一个常微分方程组，大约有100个左右要estimate的parameters
10，000个equations。equations>>parameters
所以考虑用Least Square Fit求近似解
用什么软件，什么函数可以解决
我对数值计算没有什么经验，新手问题，希望大家不吝赐教，谢谢

c*a2007-06-08 07:06

2 楼

I think that can be solved in this way.
let A be your equation matrix
X be your parameters, B is your result vector
so AX = B
Where A is a 10,000x100 matrix, X is a 100x1 matrix
B is a 10,000x1 matrix
we need to solve x, so we multiply A'(A transpose to both side)
we get (A'A)X=A'B
Now A'A becomes a 100x100 square matrix,so we can calculate the inverse
multiply the inverse on both side
we get X=(A'A)^(-1) A'B
That is just a illustration of the method, there should be more
efficient method to ac

【在 l***y 的大作中提到】

: 一个常微分方程组，大约有100个左右要estimate的parameters
: 10，000个equations。equations>>parameters
: 所以考虑用Least Square Fit求近似解
: 用什么软件，什么函数可以解决
: 我对数值计算没有什么经验，新手问题，希望大家不吝赐教，谢谢

p*e2007-06-08 07:06

3 楼

这样解不大对吧，这是典型的电力系统最小二乘状态估计，随便搜一下就能找到做法。

【在 c*******a 的大作中提到】

: I think that can be solved in this way.
: let A be your equation matrix
: X be your parameters, B is your result vector
: so AX = B
: Where A is a 10,000x100 matrix, X is a 100x1 matrix
: B is a 10,000x1 matrix
: we need to solve x, so we multiply A'(A transpose to both side)
: we get (A'A)X=A'B
: Now A'A becomes a 100x100 square matrix,so we can calculate the inverse
: multiply the inverse on both side

c*a2007-06-08 07:06

4 楼

where is wrong? what is your algorithm then?

【在 p*****e 的大作中提到】

: 这样解不大对吧，这是典型的电力系统最小二乘状态估计，随便搜一下就能找到做法。

p*e2007-06-08 07:06

5 楼

一般是求min((Ax-b)R(Ax-b)), R是权重对角矩阵，可以是I，然后对x求导得方程

【在 c*******a 的大作中提到】

:
: where is wrong? what is your algorithm then?

c*a2007-06-08 07:06

6 楼

take a look on least_square fit on wikipeida
you will find my method is mentioned under
Least squares and regression analysis
section
http://en.wikipedia.org/wiki/Least_squares
I don't think find a min of a complicated matrix will be easy
when you have so many parameters.
On the other hand, finding the inverse is something can be
either done easily or find some alternative way to solve.
Please show your reference

【在 p*****e 的大作中提到】

: 一般是求min((Ax-b)R(Ax-b)), R是权重对角矩阵，可以是I，然后对x求导得方程

a*n2007-06-08 07:06

7 楼

dude, pinacle is right. His argument could be seen in almost any textbook
about operations research or optimal problems. It is more refered as "best
fit problem".
But you are also a genius, you started from a incorrect point, go on a wrong
way and end up the absolutely correct result.
If A is 100000x100, and b=100000x1, generally solution to Ax=b is nothing
more than an empty set. 2nd, by multiply the A', you significantly expanded
the solution set.
However, if we choose R=I as pinacle suggested

【在 c*******a 的大作中提到】

: take a look on least_square fit on wikipeida
: you will find my method is mentioned under
: Least squares and regression analysis
: section
: http://en.wikipedia.org/wiki/Least_squares
: I don't think find a min of a complicated matrix will be easy
: when you have so many parameters.
: On the other hand, finding the inverse is something can be
: either done easily or find some alternative way to solve.
: Please show your reference