ode45 with matrix 1st order ode

Hi everyone, I'm trying to solve a first order ode in a matrix form using ode45:

where and

$Q=\begin{pmatrix} \sin(x) & 0 \\

0 & \cos(x) \\

\end{pmatrix}$ on .

Here is my code:

 clear all
z = 0, %parameter 
n = 2;
T0 = eye(n,n)   
xspan = [0 5*pi];
    opts = odeset('RelTol',1e-8,'AbsTol',1e-8);
    [Tl] = ode45(@(x,T) odeTL(x,T,z,n),xspan,T0,opts);
    [q,~] = qr(Tl);
    Tl = q;
    T0 = Tl;
    [Tl] = ode45(@(x,T) odeTL(x,T,z,n),xspan,T0,opts); 
x1 = 5*pi;
T = deval(Tl,x1);
 
 
function dTdx = odeTL(x,T,z,n)
Q = [sin(x) 0;0 cos(x)];
V = Q-z*eye(n,n);
W = T+eye(n,n);
R = T-eye(n,n);
Omg = eye(n,n)-0.5*R'*V*W;
dTdx = T*Omg;
end

As I run the code it said 'Matrix dimensions must agree' and I dont really see how the ode45 works for the matrix case?

Answer 1

[Alan Stevens]

Does this help

z = -1; %parameter Needs to be negative or T1 blows up.
n = 2;
T0 = eye(n,n); 
xspan = [0 5*pi];
opts = odeset('RelTol',1e-8,'AbsTol',1e-8);
[x,Tl] = ode15s(@(x,T) odeTL(x,T,z,n),xspan,T0,opts);
 
plot(x,Tl),grid
 
function dTdx = odeTL(x,T,z,n)
        T = reshape(T,2,2);   % T comes in as a column vector
        I = eye(n,n);
        Q = [sin(x) 0;0 cos(x)];
        V = Q-z*I;
        W = T+I;
        R = T-I;
        O = I-0.5*R'*V*W;
        dTdx = T*O;
        dTdx = dTdx(:);      % dTdx must return as a column vector
end

[Haya M]

Thank you very much Alan, so ode45 only accepts vectors and you rehape T to a vector to solve the ode. Now how can I evaluate Tl at x=5*pi? and I need to returrn Tl to be a matrix to factorize it using qr factorization?

[Alan Stevens]

Include these lines just before (or just after) the plot command

Tlat5pi = reshape(Tl(end,:),n,n);
disp(Tlat5pi)

The values, of course, will depend on the value you use for z. I chose z = -1 arbitrarily.

Incidentally, I should really have put

T = reshape(T,n,n); 

in the loop. However, if you change n to anything other than 2, you will need to specify an equivalent size matrix for Q.

[Haya M]

yes, I got it! Thank you very much Alan.