Symbolic manipulation and integration / not enough input argument

Hello, I have two symbolic functions that i want to integrate. I have converted with matlabFunction the equations into a file here reported:

function DY = Pendulum(theta,w,l)
%PENDULUM
%    DY = PENDULUM(THETA,W,L)

%    This function was generated by the Symbolic Math Toolbox version 8.2.
%    09-Oct-2018 12:30:20

DY = [w;(sin(theta).*(-9.81e2./1.0e2))./l];

Where the variables are theta and w and l is a constant. Now I want to integrate like this:

X0 = [90*pi/180 ; 1.2];
[tout,yout] = ode45(@Pendulum, [0 5], X0, 2 );

and i get the error Not enough input arguments.

Can someone please explain what is the problem?

Answer 1

[madhan ravi]

X0 = [90*pi/180 ; 1.2];

[tout,yout] = ode45(@Pendulum, [0 5], X0)

plot(tout,yout(:,1),'r' )

hold on

plot(tout,yout(:,2),'b')

function DY = Pendulum(theta,w,l)

%PENDULUM
w=1;   % error was here because w and l were not defined , I gave w and l as an example change it according to your values

l=2;
%    DY = PENDULUM(THETA,W,L)
%    This function was generated by the Symbolic Math Toolbox version 8.2.
%    09-Oct-2018 12:30:20
DY = [w;(sin(theta).*(-9.81e2./1.0e2))./l];

end

[Mauro De Francesco]

Actually my w is a variable to integrate, the equation is: theta = integral of w

[Stephan]

I dont think w (omega) is linear rising.

[madhan ravi]

Limits of integral w?

[Mauro De Francesco]

no limits on w, just the time span changes

[madhan ravi]

can you show how w is evaluated??

[Mauro De Francesco]

w is just a symbolic variable

syms m l t theta thetad th(t) th_d(t) w w_d

[Stephan]

@Madhan: The ode of pendulum is a second order ode. The system of odes given here is the converted first order sytem which is equivalent to the equation of pendulum i think.

[madhan ravi]

@stephen maybe but I’m stuck at this point .

Answer 2

[Walter Roberson]

 ode45(@(theta, w) Pendulum(theta, w, SomeValueForL)

[Mauro De Francesco]

First error typing your code:

 When the first argument to ode45 is a function handle, the tspan and y0 arguments must be
  supplied.

If I put in tspan and y0 as suggested:

@(THETA,W)PENDULUM(THETA,W,2) returns a vector of length 3, but the length of initial conditions vector is 2. The vector returned by @(THETA,W)PENDULUM(THETA,W,2) and the initial conditions vector must have the same number of elements.

[Walter Roberson]

You are copying all of w into the output and another value as well. w is length 2 so the result is length 3. You should probably be using w(2) instead of w in constructing the output

[Walter Roberson]

w = sym('w', [2 1]);
syms theta L
om_d = .... something involving w, theta, and L
Y = [w(1) ; om_d];
matlabFunction(Y, 'File', 'Pendulum', 'Vars', {theta  w L},...
  'Outputs',{'DY'});

X0 = [90*pi/180 ; 1.2];
L = 2;
[tout,yout] = ode45(@(theta, w) Pendulum(theta, w, L) , [0 5], X0);

... When I am typing on my phone in the middle of the night, I sometimes only give the outline of the change, as editing by poking with one figure is not all that productive.

Answer 3

[Stephan]

Edit

Here is another approach, inspired by Walters answer:

syms phi(t) g l
eqn = diff(phi,t,2) + g/l * sin(phi) == 0;
eqn = odeToVectorField(eqn);
Pendulum = matlabFunction(eqn, 'File', 'Pendulum','vars',{'t','Y','g','l'},'Outputs',{'DY'});

X0 = [90*pi/180, 1.2];
l = 2;
g = 9.81;

[tout,yout] = ode45(@(t,Y)Pendulum(t,Y,g,l), [0 5], X0);
plot(tout,yout(:,1),'r',tout,yout(:,2),'b')

With the resulting Pendulum.m file:

function DY = Pendulum(t,Y,g,l)
%PENDULUM
%    DY = PENDULUM(T,Y,G,L)

%    This function was generated by the Symbolic Math Toolbox version 8.2.
%    10-Oct-2018 08:21:21

DY = [Y(2);-(g.*sin(Y(1)))./l];

Theta and w are Y(1) and Y(2) here.

Best regards

Stephan

[Mauro De Francesco]

I get that, but i can't modify the function pendulum like that because I must use matlabFunction

% Create set of ODE for pendulum solution
Y = [w ; om_d];
matlabFunction(Y, 'File', 'Pendulum', 'Vars', [theta  w l],...
    'Outputs',{'DY'});

where om_d is a symbolic function and w is a symbolic variable

[Stephan]

Then you should use the answer from Walter.

[Stephan]

See my edited answer