ode45 and plotting problem

Hi everyone. I have a matlab code, but I'm not really sure how it works. I have a function file and a script file.

Function:

function dxdt = sniffer_ode(t,x,par,tu)

X = x(1);
R = x(2);

k1 = par(1);
k2 = par(2);
k3 = par(3);
k4 = par(4);

S = interp1(tu(:,1),tu(:,2),t); 

dxdt(1) = k3*S-k4*X;
dxdt(2) = k1*S-k2*X*R;

dxdt = dxdt(:); %dxdt should be column

and the script file:

%sniffer
close all

%initial conditions:
X0=0; R0=0;
x0=[X0 R0];
%parameters:
k1=1; k2=1; k3=1; k4=1;
par=[k1 k2 k3 k4];
%input:
tu=[ 0   , 0
     1   , 0
     1.01, 1
    20   , 1];

[t,x] = ode45(@sniffer_ode,[0 20],x0, [],par,tu);

plot(t,x);

I need to plot S in the same figure as X and R. Now I don't really understand what happens at

S = interp1(tu(:,1),tu(:,2),t);

and at

[t,x] = ode45(@sniffer_ode,[0 20],x0, [],par,tu);

What are the empty brackets for? Deleting them gives me an error.

If you could help me I would be very grateful :D

Answer 1

[Star Strider]

The interp1 call interpolates the ‘tu’ matrix for whatever the value of ‘t’ is at a specific step. The ODE solvers all return the independent variable (here ‘t’), so you can easily create ‘S’ in your calling program by simply recalculating it in your main script:

tu=[ 0   , 0
     1   , 0
     1.01, 1
    20   , 1];

t = 1:20;

S = interp1(tu(:,1),tu(:,2),t);

figure(1)
plot(t, S)
grid