The partial derivative of a function that includes step functions

[fahraynk]

TL;DR

a function which includes step functions, but the step function changes with time

I have this function, and I want to take the derivative. It includes a unit step function where the input changes with time. I am having a hard time taking the derivative because the derivative of the unit step is infinity. Can anyone help me?

$S(t) = \sum_{j=1}^N I(R_j(t)) a_j\\ I(R_j) = \begin{array}{cc} \{ & \begin{array}{cc} 0 & R_j < 7 \\ 1 & R_j \geq 7 \end{array} \end{array} \\ a_j \text{ is some known constant}$
$R_j(t) : t \rightarrow Z$ ($R_j$ is an integer $\in [1,10]$ which decays over time)

$R_j$ starts at $10$ and decays over time, and I can approximate its derivative as : $\frac{dR_j}{dt}= \frac{R_j(t_2) - R_j(t_1)}{t_2-t_1}$

I would like to take the derivative of $S$

$\frac{\partial S}{\partial t} = \sum_{j=1}^n (\frac{\partial I_j}{\partial R_j} \frac{\partial R_j}{\partial t})a_j$

But, I read that the derivative of a shifted Heaviside function $H(x-7)$ is the dirac delta function $\delta(x-7)$. But, this dirac delta function, as I am aware, has an infinite value at $x=7$.

But, the change in $S$ is definitely finite because $R$ decays linearlly with time, so how can $S$ be infinite? How can I get the derivative of $S$? I want to show how $S$ changes as a function of $R$

I think the unit step is just by nature not continuous... is there an approximation I can use for this function that would be continuous?

 

[andrewkirk]

You have not specified the functions $R_1,...,R_N$. They need to be fully specified before we can consider taking derivatives.

 

[RPinPA]

S is a function which takes on discrete values, and jumps instantaneously between those values. The time derivative is either 0 (between jumps) or doesn't exist (at jumps).

But, I read that the derivative of a shifted Heaviside function H(x−7) is the dirac delta function δ(x−7).

Yes, in physics we do that. It's not exactly rigorous but makes a certain kind of sense. The Heaviside function has a vertical line in it. The slope of a vertical line is infinite.

The usual approach to make a statement like the above semi-rigorous is to make a smooth approximation, for instance a function which makes the jump in finite time. And then take the limit as that finite time gets smaller and smaller.

But, this dirac delta function, as I am aware, has an infinite value at x=7.

But, the change in S is definitely finite because R decays linearlly with time, so how can S be infinite?

Sure, S is finite. But its rate of change, being a finite change divided by a time interval of 0, is infinite. Again, the slope of a vertical line is infinite.

I think the unit step is just by nature not continuous... is there an approximation I can use for this function that would be continuous?

Sure. It's just a question of what approximation would be most convenient to work with. First thing that occurs to me is something like this:
$$\hat I(R_j) = \begin{cases}0, & R_j \lt 7 \\
\frac{R_j-7}{a}, & 7 \leq R_j \lt 7 + a \\
1, & R_j \geq 7+a
\end{cases}$$

That interpolates from 0 to 1 over an interval of width $a$, with a line of slope $1/a$. This makes $\hat I$ continuous (I'm using a "hat" to indicate that this is a modified version of your indicator function). But its derivative is still not continuous. $\hat I$ is not differentiable at $7$ or $7+a$.

So my next thing to try for the interpolation is a cubic, a function of the form $f(R-7) = a_0 + a_1(R-7) + a_2(R-7)^2 + a_3(R-7)^3$ with the conditions that it is 0 and 1 at the two endpoints and has derivative 0 at those two endpoints. With a little bit of algebra that leads me to this:

$$\hat I(R_j) = \begin{cases}0, & R_j \lt 7 \\
\frac{3}{a^2}(R_j - 7)^2 - \frac{2}{a^3}(R_j-7)^3, & 7 \leq R_j \lt 7 + a \\
1, & R_j \geq 7+a
\end{cases}$$

This version goes smoothly from 0 to 1 over the interval and is differentiable everywhere. But the piecewise nature might still cause you some headaches. So the other suggestion I have is to use something like $\hat I(R_j) = \frac{1}{2} \left [1 + \tanh \left ( \frac {R_j - 7}{a} \right ) \right ]$, which becomes the step function in the limit.

In all cases, you would take the limit as $a \rightarrow 0$.

 

[fahraynk]

Thanks @RPinPA ! Great answer. If I use that hyperbolic function, its derivative is something like $\frac{1}{2\alpha} (1 -tanh^2(\frac{R_j - 7}{\alpha} ))$

Which is a function of $\alpha$.

If $R_j$ is a measurement, would it make sense to set $\alpha$ to be the average time between measurements?