MATLAB Answers

How to solve a 2nd-order ODE system with space-dependent variable with ODE45?

1 ビュー (過去 30 日間)
Ying Wu
Ying Wu 2021 年 8 月 9 日
コメント済み: Ying Wu 2021 年 8 月 13 日
Hi, I use ode45 function to solve the following 2nd-order ODE system:
where X(t) and Y(t) are function of time, and the parameter P is dependent on the first dirivative of Y(t).
I have 15 scattered points for P and dY:
dY = [0, 0.0026, 0.0082, 0.0138, 0.0194, 0.0250, 0.0306, 0.0362, 0.0418, 0.0475, 0.0532, 0.0589, 0.0647, 0.0705, 0.0763]
P = [0, 0.0423, 0.1534, 0.2336, 0.2915, 0.3628, 0.4166, 0.4587, 0.5572, 0.8002, 0.8204, 0.5559, 0.3184, 0.1405, -0.0603]
My main steps include:
  1. I use cubic spline interpolation to fit these points and obtain a piecewise function for P(dY): pp=csapi(dY,P).
  2. Like the time-dependent variable in Matlab ode45 help, I define a vector of dY and obtain the corresponding P(dY).
  3. Treat (dY,P) as inpus of odefun and use interp1 to update P with dY at each time step during the numerical computation.
The key code are shown below:
% cubic spline interpolation
csi = csapi(dY,P);
dY = [0:0.0001:0.1]';
P = fnval(csi,dy);
% ode45
[time,sol] = ode45(@(t,Y) odefun(t,Y,dY,P),[0 1000],[0,0,0.01,0);
function dYdt = odefun(t,Y,dY,P)
CCFy = interp1(ddY,P,abs(Y(4)));
% Y(1) = X
% Y(2) = X'
% Y(3) = Y
% Y(4) = Y'
dYdt = zeros(4,1);
dYdt(1) = Y(2);
dYdt(2) = 0.51*(0.03*Y(1) + 0.0025*P - 0.045*Y(4) - Y(3)) + 0.887*Y(4) + ...
1.1642*(1-1175.51*Y(1)^2)*Y(2)- 1.774*Y(1);
dYdt(3) = Y(4);
dYdt(4) = 0.03*Y(1) + 0.0025*P - 0.045*Y(4) - Y(3);
end
But the results are incorrect. Could anyone please give me some help? Thanks!

採用された回答

Alan Stevens
Alan Stevens 2021 年 8 月 9 日
More like this perhaps
dY = [0, 0.0026, 0.0082, 0.0138, 0.0194, 0.0250, 0.0306, 0.0362, 0.0418, 0.0475, 0.0532, 0.0589, 0.0647, 0.0705, 0.0763];
P = [0, 0.0423, 0.1534, 0.2336, 0.2915, 0.3628, 0.4166, 0.4587, 0.5572, 0.8002, 0.8204, 0.5559, 0.3184, 0.1405, -0.0603];
tspan = [0 1000];
ic = [0,0.01,0,0];
[time,sol] = ode45(@(t,Y) odefun(t,Y,dY,P),tspan,ic);
subplot(2,1,1)
plot(time,sol(:,1)),grid
xlabel('time'),ylabel('Y')
subplot(2,1,2)
plot(time,sol(:,3)),grid
xlabel('time'),ylabel('X')
function dYdt = odefun(~,Y,dY,P)
% Y(1) = Y
% Y(2) = Y'
% Y(3) = X
% Y(4) = X'
p = @(dydt) spline(dY,P,dydt);
dYdt = zeros(4,1);
dYdt(1) = Y(2);
dYdt(2) = 0.03*Y(3) + 0.0025*p(Y(2)) -0.045*Y(2);
dYdt(3) = Y(4);
dYdt(4) = 0.51*dYdt(2) + 0.887*Y(2) - 1.774*Y(3) + ...
1.1642*(1-1175.51*Y(3)^2)*Y(4);
end
  3 件のコメント
Ying Wu
Ying Wu 2021 年 8 月 13 日
Thanks again for your suggestion!

サインインしてコメントする。

その他の回答 (0 件)

製品


リリース

R2020b

Community Treasure Hunt

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

Start Hunting!

Translated by