Another Dirac delta-esque integral with a step function

Question

This is an extension of a previous post I made linked below

Dirac delta-esque integral over vertically shifted and scaled step function

This time, consider again the same step function $$ S(x)=\begin{cases} m_1 & x<a \\ m_2 & x\ge a \end{cases} $$ and the additional new step function $$ H(x)=\begin{cases} C & x<a \\ 0 & x\ge a \end{cases} $$ where $C$ and $a$ are real strictly positive constants.

I am now interested in the integral of the form $$ \int_{x}^{\infty}\frac{S'(t)}{S(t)}H(t)f(t)dt $$ for arbitrary continuous function $f$. Since $H$ is discontinuous at $a$, the method used in the previous post will not work. My attempt at solving this was first to express $H$ in terms of $S$ as $$ H(x)=\frac{C}{m_1-m_2}\Bigg(S(x)-m_2\Bigg) $$ then, substitution into the integral yields $$ \frac{C}{m_1-m_2}\int_x^{\infty}S'(t)\Bigg(1-\frac{m_2}{S(t)}\Bigg)f(t)dt $$ which can be split into two integrals: $$ \frac{C}{m_1-m_2}\Bigg[\int_x^{\infty}S'(t)f(t)dt-m_2\int_x^{\infty}\frac{S'(t)}{S(t)}f(t)dt\Bigg] $$ Since $S'(t)$ is a delta function, the first integral becomes $$ \int_x^{\infty}S'(t)f(t)dt=f(a)(m_2-m_1); \ \ x<a $$ and $0$ otherwise, while the second integral using the methodology from the last post becomes $$ \int_x^{\infty}\frac{S'(t)}{S(t)}f(t)dt=f(a)\int_x^{\infty}\frac{d}{dt}\ln S(t)dt=f(a)\ln\frac{m_2}{m_1}; \ \ x<a $$ Altogether, the answer I get is $$ -Cf(a)\Bigg(1+\frac{m_2}{m_1-m_2}\ln\frac{m_2}{m_1}\Bigg); \ \ x<a $$ I have tested this numerically with functions approximating arbitrarily steep step functions and the answer seems correct, but it is off be enough for me to be suspicious the discrepancy is more than just numerics and there is actually something wrong here. Is there anything glaringly wrong about the above process I used? Any help is greatly appreciated.

Answer 1

It looks fine to me; given the "right" definition of the problem. (you find the definition at the end of the post)

As stated the initial problem as posed is not well-defined. Let me explain:

Example:

Let us look (for convenience) at the simpler problem $$I= \int_{-\infty}^{\infty} u(t) u'(t) dt = \int_{-\infty}^{\infty} u(t) du(t)\tag{1}$$ with $u(t)$ the Heaviside step function. Your way of evaluating leads to the result $$I = \frac{u(t)^2}{2}\Big|_{t=-\infty}^{\infty} =\frac12\,.\tag{2}$$ However, this derivation is more than suspicious as $u'$ does not exist.

Physically (as I understand that is your applicatin), a point mass (or concentrated mass) is a limit $u_{\epsilon}(t)$ of a family of curves that point-wise converges to $u(t)$ for $\epsilon\to 0$. For example $$ u_\epsilon(t) = \frac12 \Bigl[1+ \tanh(x/\epsilon)\Bigr]\,. $$ Replacing $u$ by $u_\epsilon$ in $I$, the integral can be evaluated for any $\epsilon$ and you obtain $I= 1/2$ independent of $\epsilon$ following the argument from (1) to (2). Here, the argument is correct as the function is differentiable.

So in this sense, the result is $1/2$. However, the family of curves $u_\epsilon$ is not unique. The argument still works though, as long as we set $u_\epsilon$ and $u'_\epsilon$ in (1) for the same family of curves.

Another family of curves that converges to $u(t)$ is $$ \tilde u_\epsilon(t) = \frac12 \Bigl[1+ \tanh\bigl((x+\alpha\epsilon)/\epsilon\bigr)\Bigr]\,,$$ $\alpha$ arbitrary. You can observe that $$\int_{-\infty}^{\infty} u_\epsilon(t) du_\epsilon(t) = \int_{-\infty}^{\infty} \tilde u_\epsilon(t) d\tilde u_\epsilon(t) = \frac{1}{2}\,.$$

On the other hand, $$\int_{-\infty}^{\infty} \tilde u_\epsilon(t) du_\epsilon(t) = 1- \int_{-\infty}^{\infty} u_\epsilon(t) d\tilde u_\epsilon(t)= \frac{1}{2} \Biggl(1+\coth \alpha - \frac{\alpha}{\sinh^2 \alpha }\Biggr)\,. $$ So you see that depending on what family you put, any result between 0 and 1 is possible.

Your problem:

So we can state, in which sense you have evaluated the integral correctly. We replace the step function by a single family of curves $u_\epsilon(t)$. We then write $$S(t) = m_1 + (m_2-m_1) u_\epsilon(t)$$ and $$H(t) = C (1- u_\epsilon(t))\,,$$ where everywhere there is the same $u_\epsilon$. Given this, your derivation is valid and you can take the limit $\epsilon \to 0$ in the end, to obtain the result quoted.

If you, on the other hand, mix $u_\epsilon$ and $\tilde u_\epsilon$ (and potentially other families that converge to $u$) then essentially any result is possible.