Finding Roots between a Differential and Piecewise Equation

Hey all!

I am currently working on a project where I have to plot the roots between two of my functions with the help of fzero. I understand that I need to create a third function h(x) where h(x) = g(x) - f(x). But one of the functions is a piecewise equation. Therefore, I am not 100% sure how to find the roots between the two.

This is my piecewise function:

xp1 =@(t) ((-(8*2)/(2^2+1))*cos(t))+((8/(2^2+1))*sin(t));
x1 =@(t) (((8*2)/(2^2+1))+6)*exp(-t/2) + xp1(t);
x2 =@(t) x1(pi)*exp((pi-t)/a);

This is my other function:

fun1 =@(t) (((4+((8*2)/(2^2+1)))*exp(-2*t))+((8/(2^2+1))*(sin(t)-2*cos(t))))

I am able to plot both of them but I dont understand how to find the roots between these two. Any help would be appreciated.

[Matt J]

This is my piecewise function:

In what way is it a piecewise function? You have listed three different functions, all presumably defined for all . Or have you left the intervals on which each is defined unspecified?

Answer 1

[Matt J]

a=1;

[t_root,fval]=fzero(@(t)rootFcn(t,a), pi)

t_root = -0.1634

fval = 0

function out=rootFcn(t,a)
 xp1 =@(t) ((-(8*2)/(2^2+1))*cos(t))+((8/(2^2+1))*sin(t));
 x1 =@(t) (((8*2)/(2^2+1))+6)*exp(-t/2) + xp1(t);
 x2 =@(t) x1(pi)*exp((pi-t)/a);
 fun1 =@(t) (((4+((8*2)/(2^2+1)))*exp(-2*t))+((8/(2^2+1))*(sin(t)-2*cos(t))));
 
 out=fun1(t);
 
 if t<=pi
    out=out-x1(t);
 else
    out=out-x2(t); 
 end
 
end

[Walter Roberson]

However fzero uses logic that depends upon continuous derivatives, and will not necessarily work on piecewise functions (even continuous ones)

[Matt J]

Is that true? I don't see any indication of that in the doc.

[Paul]

The code is an m-function and doesn't appear to depend on continuous derivatives. Even this seems to come up with reasonable answer

f = @(x) sign(x);
fzero(f,1)

ans = -4.5963e-16

fzero(f,-1)

ans = 4.5963e-16

[Walter Roberson]

https://www.mathworks.com/help/matlab/ref/fzero.html#btoc6lj-17

The algorithm, created by T. Dekker, uses a combination of bisection, secant, and inverse quadratic interpolation methods

I would note that all three of those methods involve interpolation mathematics that is not valid if the derivatives are not continuous.