code for Adams Bashforth
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lamda1=0.2;
mu1=1-lamda1;
lamda2=0.3;
mu2=1-lamda2;
lamda3=0.5;
mu3=1-lamda3;
A1=lamda1+lamda2+lamda3;
A2=lamda1+lamda2+mu3;
A3=lamda1+mu2+lamda3;
A4=lamda1+mu2+mu3;
A5=mu1+lamda2+lamda3;
A6=mu1+lamda2+mu3;
A7=mu1+mu2+lamda3;
A8=mu1+mu2+mu3;
Q=[-A1 mu3 mu2 0 mu1 0 0 0;lamda3 -A2 0 mu2 0 mu1 0 0;lamda2 0 -A3 mu3 0 0 mu1 0;0 lamda2 lamda3 -A4 0 0 0 mu1;lamda1 0 0 0 -A5 mu3 mu2 0;0 lamda1 0 0 lamda3 -A6 0 mu2;0 0 lamda1 0 lamda2 0 -A7 mu3;0 0 0 lamda1 0 lamda2 lamda3 -A8];
P=[1;0;0;0;0;0;0;0];
how can i found Q*P by Adams Bashforth ?
1 件のコメント
Jan
2019 年 5 月 3 日
The question is not meaningful: All you have to do is to write Q*P, so there is no need for an integrator.
You have posted a bunch of unclear question in the forum now. Please care for adding the details which are required to understand, what you want to do.
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