Quiver with a system of ODEs.

Hello everyone.

I am trying to plot the solutions to a system of ODEs to see how it matches to the field that quiver would plot.

The system:

x' = cos(x-y) , x(0) = x0
y' = sin(x*y)  , y(0) = y0

Unsure whether I managed to translate the system into Matlab, but here is the code:

f = @(x,y) [cos(y(1)-y(2)) sin(y(1).*y(2))]

The rest of the script:

x = linspace(0,3);
y = linspace(0,3);
grid on
[X,Y] = meshgrid(x,y);

u = ???; v = ???;
quiver(X, Y, u, v, 0.9)

hold on
y0 = 1;
x0 = 2;
[T,Y] = ode45(f,[0 3],[x0; y0]);
plot(T,Y)

I don't understand how you're suppposed to use quiver. I don't quite understand what the help documentation is saying about "u" and "v". I want to see how the solutions follow the vector field produced by quiver but have no idea how to define u or v. The inital conditions y0 and x0 can be varied to see how the solutions would follow the vector field. Thanks a lot!

[darova]

You want to solve ODE or plot vector field? These are different things in this case

[Johan Bergman]

I suppose I want to do both. But I think if I did everything correctly, I managed to solve the ODE. I want to plot the vector field.

[darova]

For vector field

u = x'; v = y';

To plot the curve

[T,Y] = ode45 ...
plot(Y(:,1),Y(:,2))    % (x,y)

[Johan Bergman]

Thank you very much! I still have some questions though. What does the ":,1" or ":,2"

denote or mean. I guess its about the first and second element of the array?

[darova]

I suppose x and y depends on t (time). As you assumed above

% f = @(x,y) [cos(y(1)-y(2)) sin(y(1).*y(2))]
f = @(t,u) [cos(u(1)-u(2)) sin(u(1).*u(2))]
[T,U] = ode45 ...
% u(1) == U(:,1) == x
% u(2) == U(:,2) == y

[Johan Bergman]

Ah, yes I understand. Thanks a lot for the help!

[Adam Danz]

@darova , perhaps your comment could be copied to the answers section so Johan Bergman can accept it.

[darova]

I agree, Adam. Thanks bro

Answer 1

[darova]

For vector field

u = x'; v = y';

To plot the curve

[T,Y] = ode45 ...
plot(Y(:,1),Y(:,2))    % (x,y)

I suppose x and y depends on t (time). As you assumed above

% f = @(x,y) [cos(y(1)-y(2)) sin(y(1).*y(2))]
f = @(t,u) [cos(u(1)-u(2)) sin(u(1).*u(2))]
[T,U] = ode45 ...
% u(1) == U(:,1) == x
% u(2) == U(:,2) == y