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.

 采纳的回答

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 个评论

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.
Dear STAR
many many thanks.
It worked now.
Thanks
As always, my pleasure.
[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.
Related question in this new thread.

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