Error in ode45 (line 115) odeargumen​ts(FcnHand​lesUsed, solver_name, ode, tspan, y0, options, varargin);

I'm new to matlab and to coding in general...

I don't get the errors, I have them both in 'odefun' and in 'ode45'

how should I solve them?

function:

%odefun
function dudz = odefun(z, u)
%constants

P=1000;% carico agente
R=20;%raggio medio
L=1000;%lunghezza tubo
s=0.5;%spessore di parete
ni=0.3;%coeff poisson
E=250000;%modulo young
w=3; %pressione interna/esterna
E1=E/(1-ni^2);%modulo young dilataz laterale impedita
I=2*pi*R*s^3/(12);%momento di inerzia cilindro
k=2*pi*(E*s/R);%rigidezza molla fondazione

%linear sistem for diff eq 4th order
dudz(1) = u(2);
dudz(2) = u(3);
dudz(3) = u(4);
dudz(4)=(-1/(E1*I))*((P*u(3))+(k*u(1)));
end

error:

odefun
Not enough input arguments.

Error in odefun (line 13)
dudz(1) = u(2);

ODE :

%range:
zRange = [0,1000];
%INItIAL conditions:
u0=[0,0,0,0];

%ODE:
[zsol,usol] = ode45(@odefun,zRange,u0)

errors:

Error using odearguments (line 93)
ODEFUN must return a column vector.

Error in ode45 (line 115)
  odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);

Error in ODE_buckling_cilindro (line 17)
[zsol,usol] = ode45(@odefun,zRange,u0)
 

Answer 1

[Walter Roberson]

odefun
Not enough input arguments.

Error in odefun (line 13)
dudz(1) = u(2);

When you invoke odefun by itself instead of by calling one of the ode*() routines such as ode45(), then you need to manually pass in valid parameters to the function -- a value for z, and a vector of values for u.

It is valid to call odefun by itself, and it can be useful to do so for debugging purposes, to make sure that your function returns the values you expect and the correct datatype and size of output.

However, you would not invoke your odefun directly in order to run the ode integration: instead you would invoke the name of the file that you wrote that starts with your line

%range:

For example if you named that file run_z_ode.m then you would invoke

run_z_ode

and that would proceed to set up the initial conditions and invoke ode45() for your odefun .

Error using odearguments (line 93)
ODEFUN must return a column vector.

Your odefun has

dudz(1) = u(2);
dudz(2) = u(3);

where you did not initialize dudz before that.

When MATLAB was being designed, they had to make a design decision: if the user has a variable that is a scalar, and the user asks to assign to the second element of the variable using a single index only, then should MATLAB extend the variable into being a row vector or a column vector ? For example if you do

A(1) = 10;
A(2) = 20;

then "should" you end up with [10 20] (row vector) or "should" you end up with [10; 20] (column vector) ?

MATLAB had to choose one of those (or else had to make it an error to extend an array using a single subscript.) MATLAB choose to make it a row vector, [10 20]. So after your lines

dudz(1) = u(2);
dudz(2) = u(3);
dudz(3) = u(4);
dudz(4)=(-1/(E1*I))*((P*u(3))+(k*u(1)));

you will have a row vector dudz that is 1 x 4, as if you had typed

dudz = [u(2), u(3), u(4), (-1/(E1*I))*((P*u(3))+(k*u(1)))];

However, ode45(), and the other ode*() functions, specifically require that you return a column vector, as-if you had typed

dudz = [u(2); u(3); u(4); (-1/(E1*I))*((P*u(3))+(k*u(1)))];

You have a series of options: you can do any of the following:

%A:
dudz = zeros(4,1);
dudz(1) = u(2);
dudz(2) = u(3);
dudz(3) = u(4);
dudz(4)=(-1/(E1*I))*((P*u(3))+(k*u(1)));

%or B:
dudz(1) = u(2);
dudz(2) = u(3);
dudz(3) = u(4);
dudz(4)=(-1/(E1*I))*((P*u(3))+(k*u(1)));
dudz = dudz.';

%or C:
dudz(1) = u(2);
dudz(2) = u(3);
dudz(3) = u(4);
dudz(4)=(-1/(E1*I))*((P*u(3))+(k*u(1)));
dudz = dudz(:);

%or D:
dudz(1,1) = u(2);
dudz(2,1) = u(3);
dudz(3,1) = u(4);
dudz(4,1)=(-1/(E1*I))*((P*u(3))+(k*u(1)));

%or E:
dudz = [u(2); u(3); u(4); (-1/(E1*I))*((P*u(3))+(k*u(1)))];

[federico midei]

thank you wery much for the explanation. I've corrected the 'odefun' row-column vector problems and renamed the file as 'prova.m now I have:

buckling_cilindro.m :

P=1000;% carico agente
R=20;%raggio medio
L=1000;%lunghezza tubo
s=0.5;%spessore di parete
ni=0.3;%coeff poisson
E=250000;%modulo young
w=3; %pressione interna/esterna



%equazioni derivanti dall'equilibrio radiale:
E1=E/(1-ni^2);%modulo young dilataz laterale impedita
I=2*pi*R*s^3/(12);%momento di inerzia cilindro
k=2*pi*(E*s/R);%rigidezza molla fondazione

prova.m :

% eq diff con w=0, P
buckling_cilindro;

function dudz = odefun (z,u)

dudz(1) = u(2);
dudz(2) = u(3);
dudz(3) = u(4);
dudz(4)=(-1/(E1*I))*((P*u(3))+(k*u(1)));
dudz = dudz(:);
end

ODE_buckling_cilindro.m :

%ODE:

%range:
L=1000;%tube length
zRange = [0,L];

%initial conditions:

u0=[0;0;0;0];% da modificare


%ODE w=0:
[zsol,usol] = ode45(@prova,zRange,u0)

but as well i get this error for 'ODE_buckling_cilindro.m' :

ODE_buckling_cilindro
Attempt to execute SCRIPT prova as a function:
C:\Users\asus\Dropbox\Il mio PC (LAPTOP-MJCI73BH)\Documents\MATLAB\prova.m

Error in odearguments (line 90)
f0 = feval(ode,t0,y0,args{:});   % ODE15I sets args{1} to yp0.

Error in ode45 (line 115)
  odearguments(FcnHandlesUsed, solver_name, ode, tspan, y0, options, varargin);

Error in ODE_buckling_cilindro (line 13)
[zsol,usol] = ode45(@prova,zRange,u0)

I've seen tutorials about ode45 but I don't get where is wrong the code.

sorry for keep asking, hope to don't annoy you too much :)

[Walter Roberson]

When you have

buckling_cilindro;
function dudz = odefun (z,u)

then that does not mean to execute buckling_cilindro and then import all of the variables it defines into odefun.

What it means instead is that buckling_clindro should be run, and then a private function odefun should be defined (but not executed.)

Then when you request @prova in ode45, ode45 will try to execute prova as a function, passing in z and u, and that will fail because prova is a script instead of a function.

To import variables into a function, see

http://www.mathworks.com/help/matlab/math/parameterizing-functions.html

or

http://matlab.wikia.com/wiki/FAQ#How_can_I_share_data_between_callback_functions_in_my_GUI.28s.29.3F

[federico midei]

you mean like this?

file: prova.m

 %eq diff with w=0 and P
function dudz = prova1 (z,u)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
P=10000;
R=20;
L=1000;
s=0.005;
ni=0.3;
E=250000;
w=0.04; 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
E1=E/(1-ni^2);
I=2*pi*R*s^3/(12);
k=2*pi*(E*s/R);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
dudz(1) = u(2);
dudz(2) = u(3);
dudz(3) = u(4);
dudz(4)=(-1/(E1*I))*((P*u(3))+(k*u(1)));
dudz = dudz(:);
end

file: ode_prova.m :

%ODE:
%range:
L=1000;%tube length
zRange = [0,L];
%initial conditions:
u0=[0; 0; 0; 0]

%ODE
[zsol,usol] = ode45(@prova,zRange,u0);

running the function 'prova.m', I still get the 'no enough input arguments' error, but the ode_prova.m runs.

I'm not sure this code is correct and solves the 4th order ode by the way...Do you agree?

[Walter Roberson]

What is your reason for trying to run prova by itself without any input arguments?

[federico midei]

No reason, done it just to see if there were any errors, no problem with that. The problem infact is just that the usol is 0 with any value of the cosnstants (p, k, E1, etc..) so i think there’s something wrong in my code.

[Walter Roberson]

Look at your code again:

dudz(1) = u(2);
dudz(2) = u(3);
dudz(3) = u(4);
dudz(4)=(-1/(E1*I))*((P*u(3))+(k*u(1)));
dudz = dudz(:);

With boundary condition [0 0 0 0] then the [? 0 0 0] will be copied to be [0 0 0 ?] . With boundary condition [0 ? 0 ?] then the (P*u(3))+(k*u(1)) will be 0, and then no matter what the constants you use for p and so on, (-1/(E1*I)) * 0 is 0. Therefore for input [0 0 0 0], no matter what your constants are, you will get [0 0 0 0] output.

You need at least one non-zero boundary condition for that system to evolve away from 0.

[federico midei]

Firstofall I really appreciate your answers ( and patience 😬), thank you!

My need for this ode is to prescribe boundary conditions for u”0 =0 (second derivative) and for u”L =0 . L, is defined in ‘zRange’ .

I’ve put the “u0=[0 0 0 0]” just to match the dimensions of the vector...

Is it possibile to prescribe boundary cond. only for u” with ode45, or it is necessary another ode solver, like ‘bvp4c’?

[Walter Roberson]

ode45() and related ode*() functions are only designed to pin down the boundary conditions at one time, and to work from there. You could pin down the boundary conditions at t=L and work backwards in time from there, but ode*() does not permit you to pin down boundaries at times 0 and L and work from there to figure out what else you need.

For that kind of work, I believe you need the boundary value solvers such as bvp4c that you mention. I have never used them myself, so I do not know how to constrain only certain boundaries.

I would, however, point out two things:

1. [0 0 0 0] as your boundary values are satisfying the requirement that u"(0) = 0 and u"(L) = 0. If the trivial solution is not satisfactory then that implies that you have some other additional condition you could impose
2. Your system looks to be a relatively simple one -- all those constants and expressions with the constants fold down to just three values as far as the calculations care. You could probably work something out symbolically if you have the symbolic toolbox.

[federico midei]

I've wrote this symbolic one :

%variabili
syms P;% carico agente
syms R;%raggio medio
syms s;%spessore di parete
syms ni;%coeff poisson
syms E;%modulo young
syms w; %pressione interna/esterna

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

E1=E/(1-(ni^2));%modulo young dilataz laterale impedita
I=(2*pi*R*(s^3))/12;%momento di inerzia cilindro
k=2*pi*((E*s)/R);%rigidezza molla fondazione
Pcr=sqrt(4*k*E1*I)%carico agente critico(sol periodica cost)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%soluzione eq diff con w=0, Pcr
syms u(z);
u0=(0);
ode = diff(u,z,4)+(Pcr/(E1*I))*diff(u,z,2)+(Pcr/(E1*I))*u == 0;
usol1(z) = dsolve(ode);
pretty(usol1)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%soluzione eq diff con w>0, Pcr
syms u(z);
ode1 = diff(u,z,4)+(Pcr/(E1*I))*diff(u,z,2)+(Pcr/(E1*I))*u+(w/(E1*I));
usol2(z) = dsolve(ode1);
pretty(usol2)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%soluzione eq diff con W<0, Pcr
syms u(z);
ode2 = diff(u,z,4)+(Pcr/(E1*I))*diff(u,z,2)+(Pcr/(E1*I))*u-(w/(E1*I));
usol3(z) = dsolve(ode2);
pretty(usol3)

the problem is that I don't know how to include b.cond. and because of the symbolic expressions I can't plot the 'usol' to visualize them..

[Walter Roberson]

%variabili
syms P;% carico agente
syms R;%raggio medio
syms s;%spessore di parete
syms ni;%coeff poisson
syms E;%modulo young
syms w; %pressione interna/esterna

L = 1000;

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

E1=E/(1-(ni^2));%modulo young dilataz laterale impedita
I=(2*pi*R*(s^3))/12;%momento di inerzia cilindro
k=2*pi*((E*s)/R);%rigidezza molla fondazione
Pcr=sqrt(4*k*E1*I);%carico agente critico(sol periodica cost)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%soluzione eq diff con w=0, Pcr
syms u(z);
du = diff(u,z);
d2u = diff(du,z);
u0=(0);
cond = [d2u(0) == 0, d2u(L) == 0];
ode = diff(u,z,4)+(Pcr/(E1*I))*diff(u,z,2)+(Pcr/(E1*I))*u == 0;
usol1(z) = dsolve(ode, cond);
pretty(usol1)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%soluzione eq diff con w>0, Pcr
syms u(z);
ode1 = diff(u,z,4)+(Pcr/(E1*I))*diff(u,z,2)+(Pcr/(E1*I))*u+(w/(E1*I));
usol2(z) = dsolve(ode1, cond);
pretty(usol2)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%soluzione eq diff con W<0, Pcr
syms u(z);
ode2 = diff(u,z,4)+(Pcr/(E1*I))*diff(u,z,2)+(Pcr/(E1*I))*u-(w/(E1*I));
usol3(z) = dsolve(ode2, cond);
pretty(usol3)

These will still have two boundary conditions unresolved; if you know more information then you can put the restrictions into cond