1
$\begingroup$

I am wondering what is the derivative of the following function with respect to $x(t)$ in sense of distributions. $$ I\left(\int_0^t x(\tau)d\tau \leq c\right) $$ where $I$ is the indicator function and $c$ is a constant.

$\endgroup$
1
$\begingroup$

I don't think one can interpret such a derivative in the sense of distributions. Distributions generalize locally integrable functions; they require an underlying measure. The space of functions $x(t)$ is infinite-dimensional and does not carry a natural measure. A functional on this space is not a distribution, and neither is its derivative.

Formally, replacing $x(\tau)$ with $x(\tau)+\epsilon \phi(\tau)$ and taking the derivative with respect to $\epsilon$ at $\epsilon=0$ taking the difference of indicator functions, and passing to the limit we arrive at something that can be written as $$ \delta\left(\int_0^t x(\tau)\,d\tau = c\right)\,\operatorname{sign} \phi \tag1$$ The meaning of (1) is that integrating $$ \delta\left(\int_0^t (x(\tau)+\epsilon\phi(\tau))\,d\tau = c\right)\,\operatorname{sign} \phi $$ over $\epsilon \in (a,b)$ gives $$ I\left(\int_0^t (x(\tau)+b\phi(\tau))\,d\tau \ge c\right)- I\left(\int_0^t (x(\tau)+a\phi(\tau))\,d\tau \ge c\right)$$

$\endgroup$

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.