Extra independent component in ode integration affects other components

1 回表示 (過去 30 日間)
Mitsu
Mitsu 2020 年 4 月 6 日
コメント済み: Mitsu 2020 年 4 月 7 日
I am trying to understand how the ode functions in MATLAB work.
In this case, I am running an ode45 or ode113 with a state that contains 6 components. Secondly, I run the exact same problem, but in the function that contains the linear integration, I am adding a 7th component. This component ("extra_parameter" in the code below) is not included in any equation of the other 6 components.
Nonetheless, the resulting trajectories end up diverging from each other given enough time. This can be seen in the plot below the code. In this plot, the 7th component is also not represented in anyway.
Since this extra component is not included in the computation of the rest of the components, why does it affect their values throughout the integration?
Here's the minimum working example:
x0=[0.8 0.3 0.1 0 0.2 0]; % initial state
mu = 1.21506683e-2; % some constant I pass to the integration functions
tspan = linspace(0,200,10000); % integration time
options = odeset('RelTol',1e-12,'AbsTol',1e-12);
% Integration
[~,y2] = ode45(@(t,y) solver1(t,y,mu),tspan,x0,options);
[~,y1] = ode45(@(t,y) solver2(t,y,mu,1),tspan,[x0 0],options); % initial value of the 7th component is 0 in this example
% Plot
figure; hold on;
plot(y2(:,1),y2(:,2),'b','LineWidth',1);
plot(y1(:,1),y1(:,2),'r--','LineWidth',1);
hold off;
% Integration functions
function dx = solver1(~,x,mu)
% Distance
r1 = sqrt((x(1)+mu)^2+x(2)^2+x(3)^2);
r2 = sqrt((x(1)+mu-1)^2+x(2)^2+x(3)^2);
% State
dx = [ x(4)
x(5)
x(6)
x(1) + 2*x(5) - (1-mu)*(x(1)+mu)/r1^3 - mu*(x(1)-1+mu)/r2^3
x(2) - 2*x(4) - (1-mu)*x(2)/r1^3 - mu*x(2)/r2^3
0 - (1-mu)*x(3)/r1^3 - mu*x(3)/r2^3];
end
function dx = solver2(~,x,mu,a)
% Distance (same as in solver 1)
r1 = sqrt((x(1)+mu)^2+x(2)^2+x(3)^2);
r2 = sqrt((x(1)+mu-1)^2+x(2)^2+x(3)^2);
% State (same as in solver 1, with an additional 7th component)
dx = [ x(4)
x(5)
x(6)
x(1) + 2*x(5) - (1-mu)*(x(1)+mu)/(r1^3) - mu*(x(1)-1+mu)/(r2^3)
x(2) - 2*x(4) - (1-mu)*(x(2))/(r1^3) - mu*(x(2))/(r2^3)
0 - (1-mu)*(x(3))/(r1^3) - mu*(x(3))/(r2^3);
a]; % extra parameter; so that a_new = a_previous + a * time throughout the integration
end
Plot:
  5 件のコメント
darova
darova 2020 年 4 月 6 日
Indeed, there is a difference. ode45 has some adaptive algorithms for choosing step of integration (dt). Maybe it chooses slightly different steps for these functions
I can't explain it honestly. It's weird. Maybe this is some of those cases when people say "tolerance" or "machine precision"
Mitsu
Mitsu 2020 年 4 月 7 日
Seeing James' answer below, I understand that it is in the nature of adaptive step ode solvers. I tried other odes and obtained similar results. At least it's good to now know this and be aware of that when using the output of any integration. Thank you, darova!

サインインしてコメントする。

採用された回答

James Tursa
James Tursa 2020 年 4 月 6 日
The integrators do not internally step each element separately. They still step the entire state vector as a whole. So even though the derivative equations are independent, they will all affect the overall internal stepping that the integrator does to arrive at a solution. And if you change the stepping, you will change the result for all elements, even if only slightly.
  3 件のコメント
James Tursa
James Tursa 2020 年 4 月 7 日
編集済み: James Tursa 2020 年 4 月 7 日
The ode integrators do not give the user the option of setting the step sizes directly. All of the step sizing is adaptive and done internally based on the tolerances in effect ('RelTol' and 'AbsTol'). Slight changes in the conditions of the integration should be expected to produce slight changes in the results, but within the tolerances. For long duration integrations, all methods may eventually diverge from each other. This is particularly true of "orbit" type of ode's, where the rectangular coordinate systems used for the "circular'ish" trajectories tend to produce systematic integration errors that build up over time (often downtrack errors). To get consistency for your ode's being solved, tighter tolerances or different / higher order methods may be appropriate, but you will still get some divergence.
Mitsu
Mitsu 2020 年 4 月 7 日
I see, thank you!

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeOrdinary Differential Equations についてさらに検索

タグ

製品


リリース

R2018b

Community Treasure Hunt

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

Start Hunting!

Translated by