differential equation with delta and heaviside functions

Question

Suppose we have a differential equation given by: $$ \frac{d}{dx}f(x) = g\big(c.H(x)\big)\delta(x) $$ where $H(x)=1_{x\ge0}$ is the Heaviside step function, $c$ is a constant and $g$ is a function (for example a polynomial). I was trying to solve it by integrating from zero to a given point $t$: $$ \int_0^tdx\frac{d}{dx}f(x) = \int_0^tg\big(c.H(x)\big)\delta(x) $$ left hand side is well defined, but right hand side is subtle. Can we define this integral properly? for example, put $g(x)=x$, then, right hand side is: $$ c\int_0^tH(x)\delta(x) $$ Intuitively, it make sense to me that this integral is equals to $H(t)$. For $t>0$, we have: $$ c\int_0^tH(x)\delta(x) = c\int_{-\infty}^{+\infty} \chi_{[0,t]} \delta(x) = 1$$ where $\chi_A$ is the indicator function in the set $A$. and for $t<0$ this is zero. I'm asking if it is correct or there are some mistakes in my reasoning. I know that this is not regorous, but and would like to know is possible to define properly all these steps I've written

Answer 1

Set $g(x)=x$ and $c=1$. There is not straightforward answer. Unfortunately there is no clear way of multiplying $H$ and $\delta$. If you consider them as being distributions it is simply not possible to take the product: this is essentially because $\chi_{[0,\infty)}$ and $\chi_{(0,\infty)}$ are the same distribution, because they are equal almost everywhere (that is, $H(0)$ is not well-defined).

Theoretically, in a different setting, it is still possible to define $\int_0^t H(x)d\delta$ and $H\cdot \delta$ with some care: $\delta$ is a measure on $\mathbb R$ with respect to (say) Borel’s $\sigma$-algebra, $H$ is a measurable function, so the integral above is well-defined… Except that now it really matters if you are multiplying the integrand by $\chi_{[0,t]}$ or $\chi_{(0,t)}$, and it really matters if $H(0)=0$ or $H(1)=1$, because now the origin has positive measure. If $H(0)=1$ (as your definition seems to suggest), then $H\cdot\delta=\delta$; if $H(0)=0$, then $H\cdot\delta=0$. This setting is a little messy and requires extreme care (if you want to dig more, you may find this post useful). Moreover, in this setting somehow we fall outside the realm of distributions, so even if we manage to define $H\cdot\delta$, it is not immediately clear in which sense the ODE above is supposed to be satisfied.

So, there is no clear way in which the above ODE can be treated. Morally speaking, it really depends on how you define (or you want to define) $H$ and $H\cdot\delta$, and in which sense the differential equation is satisfied. If you are a “distributionalist” and you don’t want to give a meaning to $H(0)$ because $H$ is only defined a.e., then $H\cdot\delta$ is simply not well-defined. If you still want to define $H\cdot\delta$, then you need to say in which sense it is defined (what I wrote in the second paragraph gives some ideas on how to give a meaning to all the objects involved; it is not the unique way: for instance, an arguably natural choice to make would be $H\cdot \delta=\delta/2$ by symmetric considerations).

It seems to me this is one of the cases in which you need first to understand what kind of phenomenon your ODE models, then deduce what the mathematical properties of the ODE must be, and then define all the mathematical setting rigorously in such a way that it matches your intuition.