2nd order ODE with time dependent parameters
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I wanna solve a ODE question with matlab The question is
θ''(t) + 2θ'(t) + w* sin θ(t) = τ (t) , τ (t) = 5(1 − exp(−5t))
θ(0) = −π/6 and θ'(0) = 0
w = 10 when 0 ≤ t ≤ 5 or 8 otherwise
t = [0 10]
I wrote the code refer to examples in matlab ode45
ft = linspace(0,10,49); % w = f
for ft = 0:5
f = 10;
end
for ft = 5.2083:10 % w change of time range
f = 8;
end
gt = linspace(0,10,49);
g = 5.*(1-exp(-5.*gt)); % g = τ (t)
tspan = [0 10];
y0 = [-pi/6 0];
[t,y] = ode45(@(t,y) myode(t,y,ft,f,gt,g), tspan, y0);
plot(t,y)
function dydt = myode(t,y,ft,f,gt,g)
dydt = zeros(2,1);
dydt(1) = y(2);
dydt(2) = -2.*y(2)- f.*sin(y(1)) + g
f = interp1(ft,f,t); % Interpolate the data set (ft,f) at time t
g = interp1(gt,g,t); % Interpolate the data set (gt,g) at time t\
end
I got errors like this
numbers of left hand side and right hand side properties are different
오류 발생: untitled2>myode (33번 라인)
dydt(2) = -2.*y(2)- f.*sin(y(1)) + g
오류 발생: untitled2>@(t,y)myode(t,y,ft,f,gt,g) (20번 라인)
[t,y] = ode45(@(t,y) myode(t,y,ft,f,gt,g), tspan, y0);
오류 발생: odearguments (90번 라인)
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
오류 발생: ode45 (106번 라인)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);
오류 발생: untitled2 (20번 라인)
[t,y] = ode45(@(t,y) myode(t,y,ft,f,gt,g), tspan, y0);
回答（1 个）
Pratyush Roy
2021-12-1
Hi 민석 김,
While defining the myode function, one can find the interpolated function first and then compute the first and second order derivatives. The following code snippet might be helpful:
ft = linspace(0,10,49); % w = f
f = zeros(1,49);
for idx = 1:49
if ft(idx)<5.2083
f(idx) = 10;
else
f(idx) = 8;
end
end
gt = linspace(0,10,49);
g = 5.*(1-exp(-5.*gt)); % g = τ (t)
tspan = [0 10];
y0 = [-pi/6 0];
[t,y] = ode45(@(t,y) myode(t,y,ft,f,gt,g), tspan, y0);
plot(t,y)
function dydt = myode(t,y,ft,f,gt,g)
dydt = zeros(2,1);
f = interp1(ft,f,t); % Interpolate the data set (ft,f) at time t
g = interp1(gt,g,t); % Interpolate the data set (gt,g) at time t
dydt(1) = y(2);
dydt(2) = -2.*y(2)- f.*sin(y(1)) + g;
end
Hope this helps!
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2021-12-1
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