I have a matrix containing force on 5 particles in three dimensions.

F =
66.3101 180.9273 312.7912
43.8362 180.3707 304.0829
4.6838 296.7552 89.7479
39.1522 298.8534 136.5975
1.4743 5.9048 63.5243
first row represent force on 1st particle in three directions respectively.Now i want to solve the differential equation D2x =F(:,1) D2y=F(:,2) D2z=F(:,3) for each row so how should i do it?

2 件のコメント

Youssef  Khmou
Youssef Khmou 2014 年 1 月 7 日
can you provide more inputs and the nature of the solution you seek?
Sangeeta  Ydaav
Sangeeta Ydaav 2014 年 1 月 7 日
initial velocity is zero

回答 (1 件)

Youssef  Khmou
Youssef Khmou 2014 年 1 月 7 日
if you have initial velocity and position values (vx,vy,vz,x,y,z) then you need add them after each integration, otherwise, you can use integral function to get X,Y,Z as follows :
X=integral(integral(F(:,1),0.1),0.1);
Y=integral(integral(F(:,2),0.1),0.1);
Z=integral(integral(F(:,3),0.1),0.1);

2 件のコメント

Sangeeta  Ydaav
Sangeeta Ydaav 2014 年 1 月 7 日
I think this integration will give velocity .I need positions ao i think i should integrate again after adding initial velocities
Youssef  Khmou
Youssef Khmou 2014 年 1 月 7 日
In this case yes :
VX=integral(F(:,1),0.1)+VX0;
X=integral(VX,0.1);

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2014 年 1 月 7 日

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2021 年 8 月 20 日

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