How to solve system of 2nd order differential equations using ode45
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Ryan Bowman
am 6 Dez. 2018
Beantwortet: Bob
am 14 Feb. 2023
I have three 2nd order differential equations with my initial conditions and I'm trying to use the ode45 function in matlab to solve this. I wish to get the solution where my output is x,y,z position vs. time plot(2nd derivative) as well as a dx,dy,dz velocity vs. time plot. I get multiple errors and I'm not sure how to fix it. Here is my code:
%Clohessy-Wiltshire Equations
% d2x = 2*n*dy + 3*(n^2)*x;
% d2y = -2*n*dx;
% d2z = (-n^2)*z;
%
% %Initial Conditions
% x(0) = -0.066538073651029; %km
% y(0) = 0.186268907590665; %km
% z(0) = 0.000003725378152; %km
% dx(0) = -0.000052436200437; %km/s
% dy(0) = 0.000154811363681; %km/s
% dz(0) = 0.000210975508077; %km/s
%Constants
a = 6793.137; %km
mu = 398600.5; %km^3/s^2
n = sqrt(mu/a^3);
t0 = 0; %seconds
tf = 5400; %seconds
initial_condition = [x(0) y(0) z(0)]
[T,position] = ode45(@(t,position)Clohessy_Wiltshire(t,x,y,z),[t0 tf],initial_condition);
figure(1), hold on, plot(T,position(:,1),'b'), plot(T,position(:,2), 'r'), plot(T,position(:,3), 'g')
title('Position(X,Y,Z) vs. Time')
ylabel('Position(X,Y,Z)(km)')
xlabel('Time(s)')
figure(2), hold on, plot(T,position(:,4),'b'), plot(T,position(:,5), 'r'), plot(T,position(:,6), 'g')
title('Velocity(dX,dY,dZ) vs. Time')
ylabel('Velocity(dX,dY,dZ)')
xlabel('Time(s)')
function position = Clohessy_Wiltshire(~,x,y,z,dx,dy,dz,n)
x(0) = -0.066538073651029;
dx(0) = -0.000052436200437;
dx(2) = 2*n*dy;
y(0) = 0.186268907590665;
dy(0) = 0.000154811363681;
dy(2) = -2*n*dx;
z(0) = 0.000003725378152;
dz(0) = 0.000210975508077;
dz(2) = -n^2*z;
end
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madhan ravi
am 6 Dez. 2018
Bearbeitet: madhan ravi
am 6 Dez. 2018
Here you go!
syms x(t) y(t) z(t)
%Clohessy-Wiltshire Equations
% d2x = 2*n*dy + 3*(n^2)*x;
% d2y = -2*n*dx;
% d2z = (-n^2)*z;
%Constants
a = 6793.137; %km
mu = 398600.5; %km^3/s^2
n = sqrt(mu/a^3);
t0 = 0; %seconds
tf = 5400; %seconds
dx=diff(x,t);
dy=diff(y,t);
dz=diff(z,t);
%Initial Conditions
c1 = -0.066538073651029; %km
c2 =0.186268907590665; %km
c3 =0.000003725378152; %km
c4 = -0.000052436200437; %km/s
c5 =0.000154811363681; %km/s
c6 = 0.000210975508077; %km/s
y0 = [c1 c2 c3 c4 c5 c6];
eq1 = diff(x,2) == 2*n*dy + 3*(n^2)*x;
eq2 = diff(y,2) == -2*n*dx;
eq3 = diff(z,2) == (-n^2)*z;
vars = [x(t); y(t); z(t)]
V = odeToVectorField([eq1,eq2,eq3])
M = matlabFunction(V,'vars', {'t','Y'});
interval = [t0 tf]; %time interval
ySol = ode45(M,interval,y0);
tValues = linspace(interval(1),interval(2),1000);
yValues = deval(ySol,tValues,1); %number 1 denotes first position likewise you can mention 2 to 6 for the next 5 positions
plot(tValues,yValues) %if you want to plot position vs time just swap here
The graph of the first position looks like below:
5 Kommentare
James VanderVeen
am 13 Apr. 2021
You specifiy the variable:
vars = [x(t); y(t); z(t)]
But you never use it. Is there a reason for inserting this into the code?
Shantanu Chhaparia
am 20 Feb. 2022
hey! are there some examples (of system of higher order differential equation) on matlab site? If yes, can you please share the link. I was unable to locate them and I had certain doubts which might get cleared by looking over those. Thanks.
Weitere Antworten (1)
Bob
am 14 Feb. 2023
Hopefully, it is valid
% define n, where Earth GM : μ = 398600.442 km³/s²
n = sqrt(398600.442e9/earthRadius^3) ; % note: earthRadius < a
% and matrices A and B
A = [0 2 0; -2 0 0; 0 0 0].* n ;
B = [3 0 0; 0 0 0; 0 0 -1].* n^2 ;
% Define symbolic variable t and vector u(t) ≡ [x; ẋ]
syms t ;
syms u(t) [6,1] matrix ;
% Specify initial value / start position
s = [-66.538073651029;186.268907590665;0.003725378152;... % m
-0.052436200437; 0.154811363681;0.210975508077] ; % m/s
% Define ODE function...
M = @(t,u)[u(4:6); A*u(4:6) + B*u(1:3)] ;
% ...and solve by ode45 on (0;5400] time interval, s = u(0)
z = ode45(M,[0 5400],s) ;
r = 0:54:5400 ; % points range to plot the results
plot(r,deval(z,r,[1:3])); % distance [m] vs time
plot(r,deval(z,r,[4:6])); % velocity [m/s] vs time
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