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As the title suggest, I am looking to understand which is the result of such operation. In fact, I am willing to find the root (numerically) of:

$$ \sqrt{(f(x)H(f(x)))^2+(g(x)H(g(x)))^2}-K=0 $$

The derivative of such thing is:

$$ \frac{1}{\sqrt{H(f(x))^2+H(g(x))^2}}\left[f(x) H(f(x)) \left(\frac{\partial f(x)}{\partial x}H(f(x))+f(x) \delta(x) \frac{\partial f(x)}{\partial x} \right)+ g(x) H(g(x)) \left(\frac{\partial g(x)}{\partial x}H(g(x))+g(x) \delta(x) \frac{\partial g(x)}{\partial x} \right)\right] $$

Now, the terms like $f^2(x) H(f(x)) \delta(x)$, for $x=0$ are not defined (or at least is what my engineering head tells me, being infinity times zero undetermined).

On the other side, if I approximate

$$ H(x) \approx \bar{H}(x,\epsilon)= \left\{ \begin{array}{l} 0 \quad x \le -\epsilon \\ \frac{x + \epsilon}{2 \epsilon} \quad -\epsilon<x<\epsilon \\ 1 \quad x \ge \epsilon \\ \end{array} \right. \quad \text{ and } \quad \delta(x) \approx \bar{\delta}(x,\epsilon)= \left\{ \begin{array}{l} 0 \quad x \le -\epsilon \\ \frac{1}{2 \epsilon} \quad -\epsilon<x<\epsilon \\ 0 \quad x \ge \epsilon \\ \end{array} \right. $$

Then I have

$$ \lim_{\epsilon \rightarrow 0} \bar{H}(0,\epsilon)\bar{\delta}(0,\epsilon)= \infty $$

However, this does not convince me at all because, as I saw here Relation between Heaviside step function to Dirac Delta function , the Dirac delta makes sense only in a integral sense.

How can I solve this ambiguity?

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  • $\begingroup$ The Dirac delta can't be multiplied with something (essentially) discontinuous at $0$. $\endgroup$ – Ian Oct 20 '15 at 21:06
  • $\begingroup$ Good observation.. but also the Dirac delta is (essentially) discontinuous at $0$... or not? $\endgroup$ – L. B. Oct 20 '15 at 21:09
  • $\begingroup$ The Dirac delta itself is (essentially) discontinuous at $0$, but the problem is that the Heaviside is also (essentially) discontinuous at $0$. Thus "in reality" $H(0)$ makes no proper sense (even though we often give it a value anyway). Anyway, when you smear out the Heaviside into something which continuously but rapidly rises, and correspondingly smear out the Dirac delta into something supported on a small interval, you get a result that you can make sense of. $\endgroup$ – Ian Oct 20 '15 at 21:09
  • $\begingroup$ (Cont.) In technical language, this is mollification. But the final result will depend on how you mollify, which is bad for a lot of reasons. Maybe in your engineering application there is some phenomenon that restricts the mollifications you can take, for instance to symmetric mollifiers. This can remove the ambiguity. But in mathematical generality there is no way to remove it. $\endgroup$ – Ian Oct 20 '15 at 21:12
  • $\begingroup$ I agree with you.. however, if you take the limit to $0^+$, that is I come from the right side of zero, then the Heaviside function is defined and it is 1... that is why I tried to approximate the problem.. By the way, what do you mean by "essentially"? $\endgroup$ – L. B. Oct 20 '15 at 21:15

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