-1
$\begingroup$

Here's my piecewise function in which $\eta$ is a parameter: \begin{align} h(\eta)=\begin{cases} f(x) & x\leq g(\eta) \\ 0 & x>g(\eta) \end{cases} \end{align} where $x$ is a positive continuous strictly increasing function of $\eta$, and $f,g,h,$ are positive continuous strictly increasing functions of their arguments. I wish to find the extremum of function $h$, for which I need $dh/d\eta$. I don't know how to even proceed with this one. This equation actually comes from an optimization problem modeling a physical system. Any help is appreciated. Thanks in advance.

$\endgroup$
1
$\begingroup$

Since it is given that $h$ is positive, the inequality $x>g(\eta)$ can never hold. Thus $h(\eta)=f(x)$. Does this help?

$\endgroup$
  • $\begingroup$ So what you are saying is $\frac{dh}{d\eta}=\frac{df}{dx}\frac{dx}{d\eta}$? $\endgroup$ – Deep Apr 25 '17 at 8:58
  • $\begingroup$ Yes, this should be the case. But if you just want to find the extremum of $h$, note that $f$ is strictly increasing in $x$, i.e. to increase $f$ is to make $x$ large. But as $x$ is always strictly increasing in $\eta$, I think $h$ can only be maximized (resp. minimized) when $\eta \to +\infty$ (resp. $-\infty$). In other words, there's no such local extremum. $\endgroup$ – delt31 Apr 25 '17 at 9:05
  • $\begingroup$ But it all depends on the fact that $h$ is positive. $\endgroup$ – delt31 Apr 25 '17 at 9:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.