How to solve a set of matrix differential equations given below in Matlab?

Consider

for t(0,30)

where

C = 1/2 ,

 [ 10    2     -5 ;...
   2    -10     0 ;...
  -5     0     -30 ] 

(i am unable to write the matrix using latex editor for some reason, its a 3x3 matrix)

and

 [sin(t)   0    0     ;...
  0        1    0     ;...
  0        0    cos(t)] 

and plot the matrix components y11(t),y12(t), y13(t),y22(t),y23(t) and y33(t)

Answer 1

[Star Strider]

Try this:

y0 =  [ 10    2     -5;    2    -10     0;  -5     0     -30 ];
C = 0.5;
f = @(t) [sin(t)   0    0 ;  0     1    0 ;  0        0    cos(t)];
t = linspace(0, 30, 75);
for k = 1:numel(t)
    y(:,:,k) = (expm(C*t(k))+y0) * f(t(k));
    yp(k,:) = [y(1,1,k) y(1,2,k) y(1,3,k) y(2,2,k) y(2,3,k) y(3,3,k)];
end

figure
plot(t, yp)
grid
legend('y_{1,1}', 'y_{1,2}', 'y_{1,3}', 'y_{2,2}', 'y_{2,3}', 'y_{3,3}', 'Location', 'SW')

with:

Experiment to get the result you want.

[Sajjan Gundmi Satish]

How is this computing the dy/dt ? Wouldn't it require ode function?

y0 = [10 2 -5 2 -10 0 -5 0 -30]';
tspan = [0,30];
C=1/2;
for ii =1:9
    [t,y] = ode45(@fode1,tspan,y0(ii));
    figure(ii)
    plot(t,y)
end
function dy = fode1(t,y)
C = 1/2;
% f = [sin(t) 0 0 0 1 0 0 0 cos(t)]';
dy = -C*y+sin(t);

The above gives me all the y values but only at the first f(t).

How should i go about it for all the values of f(t)?

[Star Strider]

My code does not compute . It integrates the differential equation using the matrix exponential function, and plots the result.

This is a linear matrix differential equation. You can certainly use ode45 if you want, however an efficient result is to use the expm function to compute the matrix exponential. (This goes back to the Cayley-Hamilton theorem and functions of a square matrix. Most linear algebra texts and all modern control texts discuss this in some detail.)

You would certainly need ode45 if your differential equations were nonlinear. That does not apply here.

A corrected version of your code (that does not use the entire matrix) will of course produce a different result:

fode1 = @(t,y) -1/2*y+sin(t);
y0 = [10 2 -5 2 -10 0 -5 0 -30]';
tspan = [0,30];
C=1/2;
for ii =1:9
    [t,y] = ode45(fode1,tspan,y0);
    plot(t,y)
end
figure
plot(t,y)
grid

This plots all the results.

I admit that I cannot code your system to use ode45 and use your initial conditions matrix. However this is straightforward using expm.