Too many output arguments ode23t
1 vue (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Param Budhraja
le 16 Oct 2020
Modifié(e) : Stephen23
le 16 Oct 2020
I am using this function
function maxh_d(t,x)
dxdt = zeros(2,1); % x(1) refers to x ,x(2) refers to p
dxdt(1)= -x(1) ; %+u
dxdt(2)= 2*x(1)*x(2);
end
in this code
[t_v,x_v] = ode23t(@(t,x) maxh_d(t,x), [0 10], [0.5 0.5]);
figure('Name','Nonlinear');
subplot(2,1,1)
plot(t_v, x_v(:,1));
I am getting this error
Error using maxh_d
Too many output arguments.
Error in mh_result>@(t,x)maxh_d(t,x) (line 1)
[t_v,x_v] = ode23t(@(t,x) maxh_d(t,x), [0 10], [0.5 0.5]);
Error in odearguments (line 90)
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode23t (line 143)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);
Error in mh_result (line 1)
[t_v,x_v] = ode23t(@(t,x) maxh_d(t,x), [0 10], [0.5 0.5]);
0 commentaires
Réponse acceptée
Bjorn Gustavsson
le 16 Oct 2020
Try to change your maxh_d function to explicitly return dxdt:
function dxdt = maxh_d(t,x)
dxdt = zeros(2,1); % x(1) refers to x ,x(2) refers to p
dxdt(1)= -x(1) ; %+u
dxdt(2)= 2*x(1)*x(2);
end
I've always written my ODE-functions so that they return the derivative. How you get that error-message I've no idea - since to my understanding 0 output arguments are one too few...
HTH
3 commentaires
Bjorn Gustavsson
le 16 Oct 2020
...Ah, perhaps I should spend some time reading and thinking before answering...
Plus de réponses (0)
Voir également
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!