- Home
- >
- Dirac Delta Function – Definition, Form, and Applications

JUMP TO TOPIC

# Dirac Delta Function – Definition, Form, and Applications

The**Dirac delta function**is an important tool to learn, especially when you’re planning to study advanced statistics, engineering, and physics concepts such as probability distributions, impulse functions, and quantum mechanics. At first glance, the Dirac delta function may appear intimidating, but once you break down the concepts, Dirac delta will help you understand how complex functions work!

**In this article, we’ll cover all the fundamental concepts and properties needed for you to understand Dirac delta functions. We’ll also relate these properties to make you appreciate them more. You’ll also get your hands on initial value problems that involve Laplace transformation and Dirac delta function.**

*The Dirac delta function is a generalized function that is best used to model behaviors similar to probability distributions and impulse graphs. We can also use the Dirac delta function to solve more complex differential equations with the help of Laplace transforms.*## What Is a Dirac Delta Function?

The Dirac delta function is an essential “function” in advanced calculus and physics (particularly, quantum mechanics). A great way to visualize what Dirac delta functions represent is by modeling a mass distribution – Dirac delta functions will exhibit similar behaviors. This means that we observe the behavior of a function at these periods:- starting when it is near zero
- its increase is suddenly drastic, along with a period of an interval, and
- when the increase slows down and eventually reaches back to zero.

### Dirac Delta Function Properties

The Dirac delta function has a wide range of properties that can help you in evaluating integrals, simplifying differential equations, and applying them to model impulse functions, along with other applications. The three main properties that you need to be aware of are shown below.**The Dirac delta function, $\delta(x – x_0)$ is equal to zero when $x$ is not equal to $x_0$.\begin{aligned} \delta(x – x_0) = 0 , \text{ when } x \neq x_0\end{aligned}Another way to interpret this is that when $x$ is equal to $x_0$, the Dirac delta function will return an infinite value.\begin{aligned} \delta(x – x_0) = \infty , \text{ when } x = x_0\end{aligned}The best way to visualize this property of the Dirac delta function is to imagine how light impulses behave: there are instances when we can no longer measure the energy emitted by the light and there are certain distances when we can. On the instances when it is near impossible to measure the energy, we just assume it’s equal to infinite.**

__Property 1:__

__Property 2:__**By integrating the Dirac delta function, we can show that the function is equal to $1$ within the allowed interval.\begin{aligned}\int_{x_0 – \epsilon}^{x_0 + \epsilon} \delta(x – x_0) \phantom{x}dx = 1, \text{ when } \epsilon >0\end{aligned}The upper and lower limits, $ x_0 – \epsilon$ and $ x_0 + \epsilon$, represent the range that covers $x$. In the context of impulse, this is the range you can observe the behavior of the function.\begin{aligned}\int_{-\infty}^{\infty} \delta(x – x_0) \phantom{x}dx = 1, \text{ when } \epsilon >0\end{aligned}We can extend this property when $\epsilon >0$ and as $\epsilon$ approaches infinity.**

__Property 3:__**We can extend the second property to account for instances when we multiply $\delta(x)$ with a function, $f(x)$.\begin{aligned}\int_{x_0 – \epsilon}^{x_0 + \epsilon} f(x)\delta(x – x_0) \phantom{x}dx = f(x_0), \text{ when } \epsilon >0\end{aligned}There are instances when $\delta(x)$ is equal to zero throughout the interval, so we use nonzero functions such as $f(x)$ evaluated at $x_0$. These three properties also highlight the significance of Dirac delta functions in normal and probability distributions.From these properties, we can also see how important Dirac delta functions are in advanced statistics, quantum mechanics, and more. Dirac delta function still has a wide range of important properties, but for now, let’s put our focus on applying Dirac delta functions and see how we can use them and Laplace transforms in solving differential equations and initial value problems.**

## How To Use Dirac Delta Function in Differential Equations?

We can use the Dirac delta function to solve differential equations by connecting our understanding of Dirac delta functions and Laplace transformations. First, let’s establish the general form of the Dirac delta function’s Laplace transform.\begin{aligned}\mathcal{L}\{\delta(x – x_0)\} &= \int_{0}^{\infty} e^{-st} \delta (x – x_0) \phantom{x}dt \\&= e^{-as}\end{aligned}Keep in mind that our conditions for Laplace transforms must be maintained, so $a >0$. Here are some examples on how we can apply this formula and to be consistent with our Laplace transform notations, we’ll use $t$ instead of $x$ inside the Dirac delta function.\begin{aligned}\mathcal{L}\{\delta(t + 6)\} &= \mathcal{L}\{\delta(t – -6)\}\\&=e^{6s} \end{aligned} | \begin{aligned}\mathcal{L}\{3\delta(t – 4)\} &= 3\mathcal{L}\{\delta(t – 4)\}\\&=3e^{-4s} \end{aligned} | \begin{aligned}\mathcal{L}\{-2\delta(t +8)\} &= -2\mathcal{L}\{\delta(t – -8)\}\\&=-2e^{8s}\end{aligned} |

\begin{aligned}\boldsymbol{f(t)}\end{aligned} | \begin{aligned}\boldsymbol{\mathcal{L}\{f(t)\} = F(s)} \end{aligned} |

\begin{aligned} y^{\prime }\end{aligned} | \begin{aligned} sF(s) – f(0), \phantom{x}s > 0\end{aligned} |

\begin{aligned} y^{\prime\prime}\end{aligned} | \begin{aligned} s^2F(s)– sf(0) – f^{\prime}(0), \phantom{x}s > 0\end{aligned} |

\begin{aligned} y^{(n) }\end{aligned} | \begin{aligned} s^nF(s) – s^{n -1}f(0)- s^{n -2}f^{\prime}(0)… -sf^{(n -2)}(0)– f^{(n -1) }(0), \phantom{x}s > 0\end{aligned} |

**Find the solution to the initial value problem, $y^{\prime \prime} – 6y^{\prime} – 16y = 4\delta(t – 8)$, where $y(0) = -4$ and $y^{\prime}(0) =8$.**

*Example 1*__Solution__When working with linear differential equations, we’ll need to take the Laplace transform of both sides of the equation. Use the formulas we’ve provided in the section shown above. Substitute the $y(0)$ and $y^{\prime}$ using our initial conditions.\begin{aligned}s^2F(s) -sf(0) -f^{\prime}(0) – 6(sF(s) – f(0)) + 16F(s) &= 4e^{-8s}\\ (s^2 – 6s – 16)F(s) – sf(0) – f^{\prime}(0)+ 6f(0) &= 4e^{-8s}\\ (s^2 – 6s – 16)F(s) +4s – 8 – 24&= 4e^{-8s}\\ (s^2 – 6s – 16)F(s) +4s – 32&= 4e^{-8s}\end{aligned}Divide both sides of the equation by $s^2 – 6s -16$ and write down the resulting rational expression as decomposed partial fractions.\begin{aligned}(s^2 – 6s – 16)F(s) &= 4e^{-8s} – 4(s – 8)\\F(s) &= \dfrac{4e^{-8s}}{(s – 2)(s + 8)} – \dfrac{4(s – 8)}{(s – 2)(s + 8)}\\&= 4e^{-8s} G(s) – H(s)\end{aligned}Let’s first evaluate the inverse Laplace transform for $G(s) = \dfrac{1}{(s – 2)(s + 8)}$ and $Hs) = \dfrac{4s- 32}{(s – 2)(s +8)}$. As we have mentioned, this is a presumed knowledge, so do head over to the link in case you need a refresher.Going back to our two Laplace transform functions, here are the resulting functions if we reverse the process to find $g(t)$ and $h(t)$:

\begin{aligned} G(s) &= \dfrac{1}{(s – 2)(s + 8)}\\g(t) &= \mathcal{L}^{-1}\{G(s)\}\\&= \dfrac{1}{10}e^{2t}- \dfrac{1}{10}e^{-8t}\end{aligned} | \begin{aligned} H(s) &= \dfrac{4s- 32}{(s – 2)(s +8)}\\h(t) &= \mathcal{L}^{-1}\{H(s)\}\\&= -\dfrac{12}{5}e^{2t} + \dfrac{32}{5}e^{-8t}\end{aligned} |