Not enough input arguments.

2022 6 14
1 个回答
更新时间：2022 6 15
2 次查看（30 天）

0 个投票

i am getting error like:
>> run_Lp
xstart =
0 0 1
Not enough input arguments.
Error in l_ext (line 11)
f(2)=-x(1).*x(3)+R*x(1)-x(2);
Error in odearguments (line 90)
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode45 (line 106)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);
Error in lya (line 19)
[T,Y] = ode45(rhs_ext_fcn,[t t+stept],x,R);
Error in run_Lp (line 9)
lk= lya(n,rhs_ext_fcn,tstart,stept,tend,xstart,R);
>>

3 个评论
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Torsten
Torsten 2022-6-14
run_LP has several inputs. Where do you supply them ?
Just writing "run_LP" in the command window will make MATLAB error.
Sam Chak
Sam Chak 2022-6-14
Guess you want to find the Lyapunov exponents of the Lorenz system?
nune pratyusha
nune pratyusha 2022-6-15
yes, I want lyapunov exponent v/s parameters plot so, i am trying it

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回答（1 个）

Jan
Jan 2022-6-14

0 个投票

Do not provide parameters as appended variables in the ODE45 call:
% In lya:
[T,Y] = ode45(rhs_ext_fcn,[t t+stept],x,R);
% ^ nope
Use an anonymous function to provide parameters:
% in run_Lp:
% Nope: rhs_ext_fcn = @l_ext;
R_max=10;
R_min=1; nk=10; tstart=0; stept=0.5; tend=20; xstart=[0 0 1]
R_step=(R_max-R_min)/nk;
R=R_min;
rhs_ext_fcn = @(x, t) l_ext(x, t, R);

6 个评论
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nune pratyusha
nune pratyusha 2022-6-15
I tried your suggestion then also error is coming like
>> run_Lp
xstart =
0 0 1
Unrecognized function or variable 'R'.
Error in run_Lp>@(x,t)l_ext(x,t,R) (line 4)
rhs_ext_fcn=@(x,t) l_ext(x,t,R);R_max=10;R_min=1;nk=10;tstart=0;stept=0.5;tend=20;xstart=[0 0 1]
Error in odearguments (line 90)
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode45 (line 106)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);
Error in lya (line 19)
[T,Y] = ode45(rhs_ext_fcn,[t t+stept],xstart);
Error in run_Lp (line 9)
lk= lya(n,rhs_ext_fcn,tstart,stept,tend,xstart);
>>
Stephen23
Stephen23 2022-6-15
@nune pratyusha: this line:
rhs_ext_fcn=@(x,t) l_ext(x,t,R);
requires R to be already defined. However in your code you define R nine lines later, which will not work.
Solution: define R before that line:
R = .. % <---- define R
rhs_ext_fcn=@(x,t) l_ext(x,t,R); % <---- before you refer to R here
nune pratyusha
nune pratyusha 2022-6-15
if i am giving R=1:0.1:20;
it showing error like
>> run_Lp
xstart =
0 0 1
Index exceeds the number of array elements. Index must not exceed 3.
Error in l_ext (line 5)
X= [x(4) x(7) x(10);
Error in run_Lp>@(t,x)l_ext(t,x,R) (line 7)
rhs_ext_fcn=@(t,x) l_ext(t,x,R);
Error in odearguments (line 90)
f0 = feval(ode,t0,y0,args{:}); % ODE15I sets args{1} to yp0.
Error in ode15s (line 152)
odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);
Error in lya (line 19)
[T,Y] = ode15s(rhs_ext_fcn,[t t+stept],xstart);
Error in run_Lp (line 10)
lk= lya(n,rhs_ext_fcn,tstart,stept,tend,xstart);
>>
Stephen23
Stephen23 2022-6-15
编辑：Stephen23 2022-6-15
@nune pratyusha: originally you defined
R=R_min;
why are you now defining it as
R=1:0.1:20;
?
In any case, the cause of the error is clear: inside L_EXT() you are trying to access elements 4-10 of array x which only has 3 elements. Trying to access non-existent array elements throws an error. This is to be expected.
You defined XSTART to have three elements, therefore the x array inside the objective function will also have three elements.
I strongly recommend that you have only one assignment per line, and align your code consistently.
nune pratyusha
nune pratyusha 2022-6-15
i changed the code but i didn't get any output
function run_Lp(n,rhs_ext_fcn,tstart,stept,tend,xstart,nk);
figure()
hold on;
n=3;
R_max=10;R_min=1;nk=10;tstart=0;stept=0.5;tend=20;xstart=[0 0 1]
R_step=(R_max-R_min)/nk;
R=1:R_step:10; R =R+R_step;
rhs_ext_fcn=@(t,x) l_ext(t,x,R);
while R<=R_max
lk= lya(n,rhs_ext_fcn,tstart,stept,tend,xstart);
R =R+R_step;
plot(R,lk);
end
Stephen23
Stephen23 2022-6-15
编辑：Stephen23 2022-6-15
@nune pratyusha: the WHILE loop never runs.
Consider your variables:
R_max =
10
R =
1.9000 2.8000 3.7000 4.6000 5.5000 6.4000 7.3000 8.2000 9.1000 10.0000 10.9000
which you use on this line:
while R<=R_max
The WHILE documentation states "An expression is true when its result is nonempty and contains only nonzero elements (logical or real numeric). Otherwise, the expression is false."
Question: are all of the expression elements you provide to WHILE true?
Answer: no (because 10.9 is greater than 10, so that last element will be false).
So your WHILE loop never runs.
I doubt that R should be a vector.
I strongly recommend that you have one assignment per line, and consistently align your code.

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