필터 지우기
필터 지우기

Runge-kutta results do not align with ODE45 solver, what am I doing wrong?

조회 수: 2 (최근 30일)
I am working on simulation the movement of three bodies with mass M, position P, velocity V, acceleration A, and force applied in it F.
I am given the initial values of position and velocity to work with, and according to Newton laws I can acquire the force and the right hand side of equation to implement Runge-Kutta and acquire the position of each body, though when I implement runge-kutta and compare the results to those I get from ODE45, I find them disimilar. What am I doing wrong?
I have attached the file of the instruction I was given and have followed.
The only two files that need to be run is ODE.m and testingrunge.m.
  댓글 수: 1
Jan
Jan 2022년 2월 17일
편집: Jan 2022년 2월 17일
Just some hints for a simplification:
% Replace
P1=[X(1),X(2),X(3)];
% by
P1 = X(1:3);
% Replace
Y_matrix=cell2mat([{X(4),X(5),X(6)};
{A1(1),A1(2),A1(3)};
{X(10),X(11),X(12)};
{A2(1),A2(2),A2(3)};
{X(16),X(17),X(18)};
{A3(1),A3(2),A3(3)}]);
Y=(Y_matrix(:));
% by
Y = [X(4:6), A1, X(10:12), A2, X(16:18), A3].';
% Replace
F=((-1)*(M*M1).*(P-P1)/((norm(P-P1).^3)))-(1*M*M2.*(P-P2)./((norm(P-P2)).^3));
% by:
F= -M*M1 * (P-P1) / norm(P-P1).^3 - M*M2 * (P-P2) / (norm(P-P2).^3);

댓글을 달려면 로그인하십시오.

채택된 답변

Jan
Jan 2022년 2월 17일
편집: Jan 2022년 2월 17일
You are using wrong initial values in the Runge Kutta method:
% X(:, 1) = RHS(0, V0); Nope!
X(:, 1) = V0;
The inital values are the initial values, not the right hand side of them.
I simplified the code to make it easier to debug:
function main
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 = [P1; V1; P2; V2; P3; V3];
h = 0.01;
t0 = 0;
tEnd = 1;
[t, X] = ode45(@RHS, t0:h:tEnd, V0);
figure;
plot(t, X);
[t2, X2] = rungekutta(h, t0, tEnd, V0);
figure;
plot(t2, X2);
end
function Y = RHS(t, X)
M1 = 1; M2 = 2; M3 = 0.5;
P1 = X(1:3);
P2 = X(7:9);
P3 = X(13:15);
A1 = Force(P1, M1, P2, M2, P3, M3) / M1;
A2 = Force(P2, M2, P1, M1, P3, M3) / M2;
A3 = Force(P3, M3, P1, M1, P2, M2) / M3;
Y = [X(4:6); A1; X(10:12); A2; X(16:18); A3];
end
function F = Force(P, M, P1, M1, P2, M2)
F = -M * M1 * (P-P1) / norm(P-P1).^3 ...
-M * M2 * (P-P2) / norm(P-P2).^3;
end
function [t, X] = rungekutta(h, t0, Tend, V0)
t = t0:h:Tend;
X = zeros(numel(V0), length(t)); % Pre-allocation
X(:, 1) = V0; % Fixed bug here!
for i = 1:length(t)-1
k_1 = h * RHS(t(i), X(:, i));
k_2 = h * RHS(t(i) + 0.5*h, X(:, i) + 0.5 * h * k_1);
k_3 = h * RHS(t(i) + h, X(:, i) - k_1 + 2 * k_2);
X(:, i+1) = X(:, i) + (k_1 + 4 * k_2 + k_3) / 6;
end
end
  댓글 수: 1
Nour Butrus
Nour Butrus 2022년 2월 17일
Thank you so much! I can't believe I was losing my mind for one day over this mistake.
Thank you a lot for simplifying the code!

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Ordinary Differential Equations에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by