error occurs in the present code

function main
clc
clear all
x=3;
options=optimset('Display','iter');
x1=fsolve(@solver,x,options);
function F=solver(x)
options=odeset('RelTol',le-8,'AbsTol',[le-8, le-8, le-8]);
[t,u]=ode45(@equation,[0,20],[4 -1 x],options);
s=length(t);
F=u(s,2);
figure(1)
plot(t,u(:,2))
hold on
end
end
function dy=equation(t,y)
dy=zeros(3,1);
dy(1)=y(2);
dy(2)=y(3);
dy(3)=y(2)^2-y(1)*y(3);
end
After running the above code following error occurs:
Not enough input arguments.
Error in fsolve (line 230)
fuser = feval(funfcn{3},x,varargin{:});
Error in (line 10)
x1=fsolve(@solver,x,options);
Caused by:
Failure in initial objective function evaluation. FSOLVE cannot continue.

 Risposta accettata

You have a typographical error in your odeset call. You typed ‘l’ (lower-case ‘L’) instead of the number 1.
Use this instead:
options=odeset('RelTol',1e-8,'AbsTol',1e-8);
and your code runs without error.
When I ran it,, the result was:
x1 =
3.732

7 Commenti

MINATI
MINATI il 23 Feb 2019
Thanks for your quick and right response.
I need a single curve
But it gives multiple graphs(13) and it is taking so much time if I change
[t,u]=ode45(@equation,[0,20],[1 -1 x],options); % bold from 4 to 1(50 curves)
any modification needed in the program?
I am not certain what you want. If you want to plot a single curve (with the optimised value for the third initial condition), just plot the last one.
Try this:
x=3;
options=optimset('Display','iter');
x1=fsolve(@solver,x,options)
function F=solver(x)
options=odeset('RelTol',1e-8,'AbsTol',1e-8);
[t,u]=ode45(@equation,[0,20],[4 -1 x],options);
s=length(t);
F=u(s,2);
% figure(1)
% plot(t,u(:,2))
% hold on
% end
% end
function dy=equation(t,y)
dy=zeros(3,1);
dy(1)=y(2);
dy(2)=y(3);
dy(3)=y(2)^2-y(1)*y(3);
end
end
function dy=equation(t,y)
dy=zeros(3,1);
dy(1)=y(2);
dy(2)=y(3);
dy(3)=y(2)^2-y(1)*y(3);
end
[t,u]=ode45(@equation,[0,20],[4 -1 x1],options);
figure(1)
plot(t,u(:,2))
Esperiment to get the result you want.
MINATI
MINATI il 23 Feb 2019
Dear STAR
many many thanks.
It worked now.
Thanks
As always, my pleasure.
MINATI
MINATI il 24 Feb 2019
[t,u]=ode45(@equation,[0,20],[4 -1 x1],options);
what is the significance of 4 in [4 -1 x1]
It is one of the initial conditions for your ‘equation’ function differential equation system.
Rik
Rik il 24 Feb 2019
Related question in this new thread.

Accedi per commentare.

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Richiesto:

il 23 Feb 2019

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Rik
il 24 Feb 2019

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