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What happens when you multiply the dirac delta function, $\delta(t)$, by some scalar k? From my understanding, $\delta(t)$=0 for t≠0, and $\delta(t)$=∞ for t=0. Based on this, does k*$\delta(t)$ simply equal to $\delta(t)$ since it doesn't change the function definition?

For example, let's say I have 2$\delta(t)$ - $\delta(t)$, will this result in 0 since the 2 simply gets "absorbed" into $\delta(t)$, basically making it $\delta(t)$ - $\delta(t)$?

Kyllua
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2 Answers2

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The dirac delta function isn't really a function, strictly speaking. It's a distribution. With the definition you've supplied, it is common to place the additional constraint that $\int_{-\infty}^{\infty} \delta(x)dx = 1$, which allows to avoid the logical paradox that you've outlined.

In particular, $k \delta(x)$ is the "function" satisfying $k \delta(0) = +\infty$ and is $0$ otherwise but now with the constraint $\int_{-\infty}^{\infty} k\delta(x)dx = k.$

infinitylord
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One of the properties of Dirac Delta generalised Function is:

$$\delta(ax) = \dfrac{1}{|a|}\delta(x) \qquad \qquad a\in\mathbb{R}\backslash\{0\}$$

In your case, writing $2\delta(x)$ just means $\delta\left(\dfrac{x}{2}\right)$, which is zero when $x \neq 0$.

Enrico M.
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  • I'm not quite sure about how I should go about subtracting two $\delta(t)$ functions with different scalars from one another. Can you help? – Kyllua Nov 05 '22 at 22:29
  • Well in your case it's just a "normal" substraction: $2\delta(x) - \delta(x) = \delta(x)$. In general, $\delta(ax) - \delta(bx)$ is left as such, if you cannot reduce it using the property above. – Enrico M. Nov 05 '22 at 22:37