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?

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

Answer 1

[Torsten]

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.

[L'O.G.]

Thanks. Could you please elaborate using this example? Here z is already a vector. To supply the derivatives in vector form, we could no longer have dz(2). It would have to be something like a cell? Specifically, I would like to know what happens if the first ode results in a scalar and the second results in a vector as mentioned given this example.

[Steven Lord]

In this sample snippet of code, x is a scalar and y is a vector. z contains the contents of both x and y.

x = 42;
y = [4; 8; 15; 16; 23; 42];
z = [x; y]

z = 7×1

42 4 8 15 16 23 42

I could have done this with indexing but concatenation works just as well.

w = zeros(numel(x)+numel(y), 1);
w(1) = x;
w(2:end) = y

w = 7×1

42 4 8 15 16 23 42

[Torsten]

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

[L'O.G.]

Thank you