Why do I keep getting the same error?

7 views (last 30 days)
Alvaro
Alvaro on 2 May 2024
Commented: Alvaro on 2 May 2024
% Reporte escrito fenómenos
function Reporte
clc;clear;format compact
tauf=2;
N=10;
Bi=0.5;
thetaI(1:N-2)=1;
epsilon=linspace(0,1,N);
[tau,FUN]=ode15s(@ProbEDO,[0,tauf],thetaI)
plot(tau,FUN)
end
function ProbEDO(tau,T)
deltaepsilon=1/(N-1);
theta(1:N)=0;
theta(2:N-1)=T;
theta(1)=[4*theta(2)-theta(3)]/3;
theta(N)=(4*theta(N-1)-theta(N-2))/(3+2*Bi*deltaepsilon);
for i=2:N-1
FUN(i-1)=(theta(i+1)-2*theta(i)+theta(i-1))/(deltaepsilon^2)+(theta(i+1)-theta(i-1))/(2*epsilon(i)*deltaepsilon);
FUN=FUN.';
return
end
end
  1 Comment
Alvaro
Alvaro on 2 May 2024
Error using Reporte>ProbEDO
Too many output arguments.
Error in odearguments (line 92)
f0 = ode(t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode15s (line 148)
odearguments(odeIsFuncHandle, odeTreatAsMFile, solver_name, ode, tspan, y0, options, varargin);
Error in Reporte (line 10)
[tau,FUN]=ode15s(@ProbEDO,[0,tauf],thetaI)

Sign in to comment.

Answers (1)

Walter Roberson
Walter Roberson on 2 May 2024
function ProbEDO(tau,T)
deltaepsilon=1/(N-1);
theta(1:N)=0;
theta(2:N-1)=T;
theta(1)=[4*theta(2)-theta(3)]/3;
theta(N)=(4*theta(N-1)-theta(N-2))/(3+2*Bi*deltaepsilon);
for i=2:N-1
FUN(i-1)=(theta(i+1)-2*theta(i)+theta(i-1))/(deltaepsilon^2)+(theta(i+1)-theta(i-1))/(2*epsilon(i)*deltaepsilon);
FUN=FUN.';
return
end
end
You calculate an array FUN, but you do not return it to the caller.
Note also that your return will happen when i = 2. It is a waste of a for loop if you return within it.
Also, N is not defined within your function.
  1 Comment
Alvaro
Alvaro on 2 May 2024
I keep getting the same error, any idea on how can I fix it?

Sign in to comment.

Categories

Find more on Programming in Help Center and File Exchange

Products


Release

R2023b

Community Treasure Hunt

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

Start Hunting!