How do I plot/solve the phase portrait for functions with a range?

For example, if I have the following differential equation x'' = k

where k is equal to -r when x>0 r when x<0

How would I solve the above using ode45.

In this case there is a constant "r" (arbitrary), how would I use ode45 to solve that?

Shown below is my attempt at trying to at least form the graph of one equation (but I can't get MATLAB to reproduce the analytical solution). I'm quite new to this so any help would be appreciated.

f = @(t,y) [y(2);-r];

hold on
for y20=[0 0.2 0.4 0.6 2]
[ts,ys] = ode45(f,[0,20],[0;y20]);
plot(ys(:,1),ys(:,2))
end
hold off

[Mischa Kim]

Hi Afthab, do you mean k=-r when t>0 or x>0?

Answer 1

[Mischa Kim]

Afthab, see this answer.

[Afthab]

Unfortunately, this one doesn't help because I already have an idea on how to use the quiver function and how to repeat for multiple initial conditions. However, I don't know how to set up the differential equation for this problem because x'' = -r for x>0 and x'' = r for x<0

[Mischa Kim]

OK. Do you mean k=-r when t>0 or x>0?

[Afthab]

x>0, so the solution should look something like this based on the hand analytical solution.

[Mischa Kim]

Gotcha. Try this:

function my_DE()
x0      = 0; 
Dx0     = 0.2;
r       = 1;

tspan   = linspace(0,2,100);
options = odeset('RelTol',1e-8,'AbsTol',1e-10);
[~,X]   = ode45(@DE, tspan,[x0; Dx0],options,r);

plot(X(:,1),X(:,2))
grid
end

function dX = DE(~,x,r)
dX = [x(2); k(x(1),r)];
end

function fval = k(x,r)
if (x < 0)
    fval = +r;
else
    fval = -r;
end
end

This needs some fine tuning.