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十点钟的咖啡

金虫 (正式写手)

[求助] iteration method for density equations_help_me 已有2人参与

hi there; I need some help with the following formulas
In the interaction picture.

     (1)

Then

   (2)

This equation can be iterated. and it is

   (3)



I can understand the eq.(2), but not the eq. (3).
Is anybody know how to get the equation (3). and why do we want to do such calculation?
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walk1997

金虫 (著名写手)

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十点钟的咖啡(leongoall代发): 金币+10, 多谢应助! 2014-06-28 19:29:21
引用回帖:
11楼: Originally posted by mshwangg at 2014-06-28 11:25:58
理论上同意你得看法
要数值求解,还得积分精度够高才行
walk有好一点的积分算法推荐一下?
我用复合辛普森积分,感觉不怎么样,速度慢不说,精度也不太满意...

我算法差不多一点都不会呀  只会用mathematica做数值积分 : (
我感觉新版本的mathematica数值积分速度已经快追上c的包了
以前是调用cernlib中的积分库(fortran下)
现在mathematica下也可以调用c写的cuba包  和自带的感觉差不多
(这个包应该也提供c和fortran的调用  cuba提供了4种算法)

很高维的积分估计一般还只能用MC (如果被积函数积分结果没真实奇点的话)
(原生的不好的话,自己产生特定的分布试试)
(粒子中有些积分带真实奇点的多维积分维的柯西主值积分)

我见过比较强大的计算方法算是:
   利用多维"复平面"上的柯西定理 结合MC做积分 (自己产生特定权重的分布)
---- 我重复不出来: (
12楼2014-06-28 12:44:00
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mshwangg

至尊木虫 (正式写手)

【答案】应助回帖


感谢参与,应助指数 +1
十点钟的咖啡: 金币+1 2014-06-27 14:50:51
What your aim is to solve equation 1.
However, the statement in eq.2 and eq3 is only the method using Euler's formula. The precision of this method is the order of h^3 (h is the step).
In many cases, the h^3 precision is not sufficient for meeting your expectation.
In my opinion, it's better to use the Runge-Kutta method or use some software such as Matlab or Mathematica
2楼2014-06-27 14:27:15
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十点钟的咖啡

金虫 (正式写手)

引用回帖:
2楼: Originally posted by mshwangg at 2014-06-27 14:27:15
What your aim is to solve equation 1.
However, the statement in eq.2 and eq3 is only the method using Euler's formula. The precision of this method is the order of h^3 (h is the step).
In many case ...

My equations are about the numerical solution. I just do not get the analytic solution (3). How can we derive the eq.(3) for (1) and (2)?
3楼2014-06-27 14:48:47
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十点钟的咖啡

金虫 (正式写手)

引用回帖:
2楼: Originally posted by mshwangg at 2014-06-27 14:27:15
What your aim is to solve equation 1.
However, the statement in eq.2 and eq3 is only the method using Euler's formula. The precision of this method is the order of h^3 (h is the step).
In many case ...

This is not the Euler's method. There is always a fact of (1/2!) when we use the second order Euler' method.

and when I try to get the second derivative of the density operator. I get:

4楼2014-06-27 15:00:09
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