ODE45 change the value of a parameter at each time step

Hello everyone,

I am trying to solve a differential equation of the form:

Assuming:

tspan =0:0.1:1;

beta =0.1;

x_0 = 0;

x_0_prime = 0;

x0 = [x_0 x_0_prime ]; %initial conditions

what I have written as VALUES is:

values=[0.6 0.8 0.9 1.4 2.5 0.6 0.7 0.8 0.9 1 2.3];

and it changes for each time step, meaning that tspan= 0 => values=0.6, at tspan 0.1 => values=0.8... and so on.

I was trying to find x using ode45, but I am having problems with chaning the VALUES for each time step.

I tried the following:

for i=1:numel(tspan)

[tspan,x]=ode45(@(tspan,x)rhs_x(tspan,x,values(i)),tspan,x0);

end

figure

plot(tspan,x(:,1))

function xdot=rhs_x(tspan,x,values)

beta=0.1;

xdot=[x(2);-2*beta*x(2)-x(1)-values];

end

This does not give correct results becasue it is just solving for x using, lets say the first value of VALUES=0.6, for all time steps, then finds x, and repeats again for the next value (but not changing VALUES for each time step as I want).

Thank you very much for your help!!

Answer 1

[Star Strider]

Try something like this —

tspan = 0:0.1:1;
values = [0.6 0.8 0.9 1.4 2.5 0.6 0.7 0.8 0.9 1 2.3];
beta = 0.1;
x_0 = 0;
x_0_prime = 0;
x0 = [x_0 x_0_prime]; %initial conditions

[tvc,xc]=ode45(@(t,x)rhs_x_c(t,x,values,tspan),tspan,x0);

figure
subplot(2,1,1)
plot3(tvc, values, xc(:,1))
grid
xlabel('t')
ylabel('values')
zlabel('x_1')
subplot(2,1,2)
plot3(tvc, values, xc(:,2))
grid
xlabel('t')
ylabel('values')
zlabel('x_2')
sgtitle('Linear')

figure
plot3(tvc, values, xc)
grid
xlabel('t')
ylabel('values')
zlabel('x')
title('Linear')
legend('x_1','x_2', 'Location','eastoutside')

 
[tvd,xd]=ode45(@(t,x)rhs_x_d(t,x,values,tspan),tspan,x0);

figure
subplot(2,1,1)
plot3(tvd, values, xd(:,1))
grid
xlabel('t')
ylabel('values')
zlabel('x_1')
subplot(2,1,2)
plot3(tvd, values, xd(:,2))
grid
xlabel('t')
ylabel('values')
zlabel('x_2')
sgtitle('Stepwise')

figure
plot3(tvd, values, xd)
grid
xlabel('t')
ylabel('values')
zlabel('x')
legend('x_1','x_2', 'Location','eastoutside')
title('Stepwise')

function xdot=rhs_x_c(t,x,values,tspan)
    value = interp1(tspan,values,t);
    beta=0.1;
    xdot=[x(2);-2*beta*x(2)-x(1)-value];
 end

function xdot=rhs_x_d(t,x,values,tspan)
    value = interp1(tspan,values,t, 'next');
    beta=0.1;
    xdot=[x(2);-2*beta*x(2)-x(1)-value];
 end

The first is the ‘classic’ way to solve such problems, because ‘t’ is essentially continuous, not discrete, so interpolation is appropriate.

Other methods of interpolation (rather than the default 'linear') are possible. Using 'next' produces the second result.

Choose the interpolation method that produces the desired result.

.