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How do you solve a coupled ODE when one of the ODE results in a vector of length 3 and the other results in a scalar of length 1?

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L'O.G.
L'O.G. 2023-10-5
编辑： Torsten 2023-10-5
For instance, in the following example that I found online, if dz(2) were actually a vector, how would you modify this?
[v z] = ode45(@myode,[0 500],[0 1]);
function dz = myode(v,z)
alpha = 0.001;
C0 = 0.3;
esp = 2;
k = 0.044;
f0 = 2.5;
dz = zeros(2,1);
dz(1) = k*C0/f0*(1-z(1)).*z(2)./(1-esp*z(1));
dz(2) = -alpha*(1+esp*z(1))./(2*z(2));
end

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Torsten
Torsten 2023-10-5
编辑：Torsten 2023-10-5
All solution components have to be aggregated in one big vector z, and also the derivatives have to be supplied in this vector form. E.g. if the unknows were composed of a vector x of length 4 and a vector y of length 7, you had to work with vectors z and dz of length 4 + 7 = 11.
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Torsten
Torsten 2023-10-5
编辑：Torsten 2023-10-5
Ran in:
Let x be a scalar and y a vector of length 2.
Let the equations be
dx/dt = x
dy1/dt = 2*y1
dy2/dt = 3*y2
with initial conditions
x(0) = 1,
y1(0) = 2,
y2(0) = 3.
Then you can set up the problem as
x0 = 1;
y10 = 2;
y20 = 3;
z0 = [x0;[y10;y20]];
tspan = [0 1];
[T,Z] = ode45(@fun,tspan,z0);
X = Z(:,1);
Y = Z(:,2:3);
figure(1)
plot(T,X)
figure(2)
plot(T,Y)
function dzdt = fun(t,z)
x = z(1);
y = z(2:3);
dxdt = x;
dydt = [2*y(1);3*y(2)];
dzdt = [dxdt;dydt];
end

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