Solving fractional differential equations in Matlab using fde12 function [closed]

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Closed 8 years ago.

function dfdt=myfun(t,x)

    dfdt = [...
        x(2);
        (1.5*((x(2))^2)*(cos(3*(x(1)))))-(((pi/2)^2) * ...
            (sin((pi*t)/2)))-(20*((x(1))-(sin((pi*t)/2)))) - ...
            ((0.5*((x(2))^2)*abs(cos(3*(x(1)))))+0.1) * ...
            sat(((x(2)-((pi/2)*cos((pi*t)/2))) + ...
            (20*(x(1)-(sin((pi*t)/2)))))/0.1)-(((abs(sin(t)))+1) * ...
            (cos(3*x(1)))*((x(2))^2))
];

sat in this equation is defined as follows:

 function f = sat(y)
     if abs(y) <= 1
         f = y;
     else
         f = sign(y);
     end

I am solving it first as an ODE using ODE45 where I define the differential equations as a vector:

   [t, x] = ode45(@myfun, [0 4], [0 pi/2])  

This works fine. But when I try to solve the same set of equations using fde12:

[T,Y] = FDE12(ALPHA,FDEFUN,T0,TFINAL,Y0,h)

Now I call it:

t0 = 0;
tfinal= 4 ;
h = 0.01;
x0 = [0 pi/2];
[t, x] = fde12(0.95, @myfun, t0,tfinal, x0,h);

(alpha is the order of fractional differentiation, e.g., 0.95)

it gives the following error:

Attempted to access x(2); index out of bounds because numel(x) = 1.

Answer 1

RTFM - or in this case: the description:

The set of initial conditions Y0 is a matrix with a number of rows equal to the size of the problem

Yet, you specify

x0 = [0 pi/2];

This has two columns. If you change it to two rows:

x0 = [0; pi/2];

It will work. (I just tried with your example).