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I am presented with a task of simulating the movement of three bodies with a constant mass using Newton's laws.
The task is very easy to follow and I have no trouble understanding it, until it comes to solving the differential equation 
using Runge-Kutta, where M is the mass of a body, 
is the second derivative of the position, namely, the acceleration and F is the force applied on M due to the two other masses.
using Runge-Kutta, where M is the mass of a body, is the second derivative of the position, namely, the acceleration and F is the force applied on M due to the two other masses.Please refer to task number 3 below
Where equation 3 is 
I am finding it particularly difficult to understand why the right hand of the equation is a column consisten of what appears to be the velocity V for each body and the acceleration 
.
I am finding it particularly difficult to understand why the right hand of the equation is a column consisten of what appears to be the velocity V for each body and the acceleration . I would highly appreciate any guidance to the interpertation of the task!
2 个评论
David Hill
2022-2-14
Show us what you have done (your code) and ask a specific question.
Nour Butrus
2022-2-14
Hey! Below are the tasks I am asked to solve, with the my code.
I am a 2nd year undergraduate student so I apologise that it is not perfect :)
Task 1
clc
clear all
%task #1
%Initial conditions
p=0.05;
P1=[0,0,0]; V1=[p,p,p];
P2=[1,0,0]; V2=[-p,p,p];
P3=[0,0,1]; V3=[-p,-p,-p];
V0=transpose([P1 V1 P2 V2 P3 V3]);
%Masses/Scalars M_i
M1=1; M2=2; M3=0.5;
Task 2
function F= Force(P,M,P1,M1,P2,M2)
F=((-1)*(M*M1).*(P-P1)/((norm(P-P1).^3)))-(1*M*M2.*(P-P2)./((norm(P-P2)).^3));
end
Task 3
function Y=RHS(t,X)
F1=Force(P1,M1,P2,M2,P3,M3);
F2=Force(P2,M2,P1,M1,P3,M3);
F3=Force(P3,M3,P1,M1,P2,M2);
Y=[{V1} {F1./M1} {V2} {F2./M2} {V3} {F3./M3}];
end
采纳的回答
Equation 3 is a 2nd order system p''=..., the RK3 is given for 1st order p'=...(see it's first difference 
) =..)
) =..)See for example equation 14 here: part1.pdf (nyu.edu)
you can see how order is reduced by introducing firts time derivatives (V's) as new variables
2 个评论
Nour Butrus
2022-2-15
Thanks a lot! So it is the case that I have to represent my initial values of 
and values of 
calculated with the intital positions in the vector Y, but am I mistaken to understand that I have to use that vector as 
to implement Runge-Kutta?
and values of calculated with the intital positions in the vector Y, but am I mistaken to understand that I have to use that vector as to implement Runge-Kutta?
AndresVar
2022-2-16
f = Y = RHS(t,X). The RHS you commented above looks good just make sure P1=X(1), V1=X(2)... etc as defined in task#1. You didn't say, but I guess velocity is constant?
% at t0
Y0 = RHS(t0,X0)
% one step later Y(t0+h)=Y1
K1=h*Y0;
K2=h*RHS(t+h/2,Y0+h/2);
K3=...
Y1 = Y0 + 1/6(K1+4*K2+K3)
% two steps later Y(t0+2h)=Y2
K1=h*Y1;
K2=...
K3=...
Y2 = Y1 + 1/6(K1+4*K2+K3)
code repeats, so just turn it into a loop for N steps
h = 0.01;
t(1) = 0;
Y(1)=RHS(t0,X0) %initial
for ii = 1:N
t(ii+1) = t(ii)+h
K1 =...
K2 =...
K3 =...
Y(ii+1) = Y(ii)+1/6(K1+4*K2+K3);
end
plot the forces vs time to see if they behave nicely... if not then you can play with the step size h
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