Finding a root with interval constraint

Hello there!

I am trying to find a point x within the time interval [t-1,t] (for some t, say t = 3) so that the function attains value zero. That is, I want to solve "Q_0 + integral(a+b*sin(c*t+d)-mu,t-1,x) = 0" for x in [t-1,t]. My code is the following

    y = fsolve(@(x) Q_0+(a-mu)*(x-t+1)-(b/c)*cos(c*x+d)+(b/c)*cos(c*(t-1)+d),0,optimset('Display','off'))

wherein (a,b,c,d) satisfy a + b*sin(c*t+d), and Q_0 and mu are constants. This code has no problem. However, the solution may sometimes be outside the time interval [t-1,t], which is not what I want.

So, my question is if there is a way to restrict the routine to find a solution that lies within [t-1,t] exactly?

Thanks!

Answer 1

[Walter Roberson]

As your x0 is a scalar (0), your x are scalar, and that implies you can use fzero() instead of fsolve(). With fzero() you can pass the interval [t-1 t] as your x0.

[Chien-Chia Huang]

It works! Thanks Walter.

[Chien-Chia Huang]

Another follow-up question:

How can I make my program remain running even if I cannot find a solution in the above nonlinear equation? As the value of mu changes, there might not have a solution in [t-1,t]. Is there any way to keep it going without stopping at error msg?

Thanks!

[Walter Roberson]

If no root is found, fzero() will not throw an error. If you use the optional output arguments then by examining the exitflag argument you can detect whether an error was encountered: the exitflag will be negative.

[Chien-Chia Huang]

Thanks, Walter. My code now goes like this (value of mu changes)

[FirstVanish,~,exitflag] = fzero(@(x) queuelength(counter_qln)+(a(j)-mu)*(x-t+1)-(b(j)/c(j))*cos(c(j)*x+d(j))+(b(j)/c(j))*cos(c(j)*(t-1)+d(j)),[t-1,t])

However, it showed the error msg

Error using fzero (line 274) The function values at the interval endpoints must differ in sign.

This is why I will need to know how to get things going without seeing the above.

[Walter Roberson]

Ah, then use try/catch