Finite difference method to solve a nonlinear eqn?
4 ビュー (過去 30 日間)
古いコメントを表示
Onur Metin Mertaslan
2022 年 3 月 25 日
コメント済み: Onur Metin Mertaslan
2022 年 3 月 25 日
Hello,
I have a second order nonlinear question and I need to solve it for different times by using finite difference method but I don't know how to start it. I am quite new in Matlab. Is there anyone who can help me or at least show me a way to do this?
thank you
6 件のコメント
Torsten
2022 年 3 月 25 日
g = 9.81;
L = 1.0;
T = 1.0;
dT = 0.01;
y_0 = pi/2;
v_0 = 0;
f = @(t,y)[y(2);-g/L*sin(y(1))];
tspan = 0:dT:T;
y0 = [y_0;v_0];
[t,y] = ode45(f,tspan,y0);
plot(t,y)
採用された回答
Torsten
2022 年 3 月 25 日
編集済み: Torsten
2022 年 3 月 25 日
g = 9.81;
L = 1.0;
T = 1.0;
dT = 0.01;
y_0 = pi/2;
v_0 = 0;
f = @(t,y)[y(2);-g/L*sin(y(1))];
tspan = 0:dT:T;
y0 = [y_0;v_0];
[t,y] = ode45(f,tspan,y0);
y_linear = v_0/sqrt(g/L)*sin(sqrt(g/L)*t) + y_0*cos(sqrt(g/L)*t);
plot(t,[y(:,1),y_linear])
3 件のコメント
Torsten
2022 年 3 月 25 日
編集済み: Torsten
2022 年 3 月 25 日
The graph won't change because the dt is not the actual stepsize of the solver, but prescribes the output times for ode45. The stepsize is chosen by the solver and computed internally - you can't influence it because it's adaptively chosen and different for each time step to meet a prescribed error tolerance.
If you want a fixed-step solver to solve your equations for which you can prescribe the stepsize, you will have to program it on your own. E.g. explicit Euler is a simple one.
その他の回答 (0 件)
参考
カテゴリ
Help Center および File Exchange で Nonlinear Control についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!