# Is this a Dirac delta function in disguise? Consider a typical detector equivalent circuit, where the detector can be seen as an ideal current generator $$I(t)$$. Since $$I = I_C + I_R$$ (currents through the capacitor and resistor), $$I_R R=V$$ and $$\frac{I_C}{C}=\frac{dV}{dt}$$, the equation of the circuit is:

$$\frac{dV}{dt}+\frac{V}{RC}=\frac{I}{C}.$$

When $$C$$ goes to $$0$$, the equation should read $$V=IR$$ as it happens in detectors with a low $$RC$$ constant. But if I wanted to send $$C$$ to zero in the solution of the differential equation, which is $$V(t)=\int_0^t \frac{I}{C}e^{\frac{t'-t}{RC}}dt'$$, to obtain the same exact result, I would be brought to think that $$e^{\frac{t'-t}{RC}}$$ is kind of double a Dirac delta function. Substituting in the integral: $$V(t)=\int_0^t \frac{I}{C}2\delta(t'-t)RCdt'=I(t)R.$$

Is it actually a delta? Is it correct to reason this way?

• Whether it is correct or not, it's much easier to rewrite your equation as $CR\dfrac{dV}{dt} + V= IR$ and not play complicated mathematical games at all. Jul 16 at 18:54

1. Yes, OP has essentially constructed a one-sided Dirac delta distribution $$\delta_{[0,\infty[}(x)~=~\lim_{\varepsilon\searrow 0} \frac{1}{\varepsilon} e^{-x/\varepsilon}$$ on the positive half line $$[0,\infty[$$ via a generalized function, so that $$\int_{[0,\infty[} \mathrm{d}x~\delta_{[0,\infty[}(x)~f(x)~=~f(0)$$ for all test functions $$f:[0,\infty[\to \mathbb{R}$$.

2. Notice that OP's distribution is different from the usual two-sided Dirac delta distribution $$\delta_{\mathbb{R}}(x)$$ on the real line $$\mathbb{R}$$, which satisfies $$\int_{\mathbb{R}} \mathrm{d}x~\delta_{\mathbb{R}}(x)~f(x)~=~f(0)$$ for all test functions $$f:\mathbb{R}\to \mathbb{R}$$.

3. This also explains why OP's last formula has a correction factor of 2.

• Would you have a reference for a more detailled analysis about the "one sided delta function" ? Thanks ! Jul 16 at 17:06
• @VincentFraticelli. Do you have a proper distribution-theoretical definition of "one sided delta function"? Jul 16 at 17:19
• I found out that it exists on this post. But, I had already asked myself the question about the currents induced on the surface of a conductor when the conductivity tends towards infinity. There is the same problem because the conductor exist only for $z>0$ Jul 16 at 17:26

### A more mathematical treatment

We have an ordinary differential equation (ODE): $$\alpha^{-1}y'(t) + y(t) = f(t).$$ In our case, $$y(t)=V(t),$$ $$\alpha=\frac{1}{RC},$$ $$f(t)=RI(t).$$

The ODE can be solved by finding a Green's function $$G_\alpha(t)$$ satisfying $$\alpha^{-1}G_\alpha'(t) + G_\alpha(t) = \delta(t)$$ and then get the solution $$y(t)$$ as $$y(t) = (G_\alpha*f)(t) = \int_{-\infty}^{\infty} G_\alpha(t-t')\,f(t')\,dt'.$$

A Green's function that only gives a response for $$t>0$$ (so that it's causal) is given by $$G_\alpha(t) = \alpha e^{-\alpha t} H(t),$$ where $$H(t)$$ is the Heaviside step function.

Distributions and convergence as such are defined by how they "act" on, or work in an integral when multiplied with, a test function $$\varphi\in C^\infty_c(\mathbb{R}).$$ Therefore, to show that $$G_\alpha(t) \to \delta(t)$$ in the sense of distributions, when $$\alpha\to\infty,$$ we need to show that $$\lim_{\alpha\to\infty} \int_{-\infty}^{\infty} G_\alpha(t)\,\varphi(t)\,dt = \int_{-\infty}^{\infty} \delta(t)\,\varphi(t)\,dt = \varphi(0)$$ for all $$\varphi\in C^\infty_c(\mathbb{R}).$$

That is easy. Using the variable change $$t=s/\alpha$$ we get $$\int_{-\infty}^{\infty} G_\alpha(t) \, \varphi(t) \, dt = \int_{-\infty}^{\infty} \alpha e^{-\alpha t} H(t) \, \varphi(t) \, dt = \int_{0}^{\infty} \alpha e^{-\alpha t} \, \varphi(t) \, dt = \int_{0}^{\infty} \alpha e^{-s} \, \varphi(s/\alpha) \, ds/\alpha \\ = \int_{0}^{\infty} e^{-s} \, \varphi(s/\alpha) \, ds \to \int_{0}^{\infty} e^{-s} \, \varphi(0) \, ds = \int_{0}^{\infty} e^{-s} \, ds \, \varphi(0) = 1 \cdot \varphi(0).$$

• Might be worth mentioning it's a linear ODE, which is necessary for solving it with a Green's function. Jul 16 at 22:38
• +1, Hopefully there is a correct mathematical answer. Jul 24 at 0:52
• @LL3.14. I would say that this is a correct mathematical answer. There are just a few things to fix to make it fully rigorous. Jul 24 at 8:35
• Oh yes, sorry if I was not clear, I was saying that your answer was a rigorous. Jul 24 at 21:49