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### Course: Differential equations>Unit 3

Lesson 2: Properties of the Laplace transform

# Dirac delta function

Introduction to the Dirac Delta Function. Created by Sal Khan.

## Want to join the conversation?

• At , I guess you could say he's building a "tau-er". Haha.
• Hi everyone, i wanted to know, why at Sal modelled the force applied with a Dirac's Delta ? Can't understand the motivation, and why this is possible...

Regards
• Consider a physical system, in which a object is at rest, and an external force is quickly applied, accelerating it to a velocity of 2 m/s. We know that this force could not have accelerated the object instantly, but for our purposes, let's assume it did. To model this, we would need a function the represents an infinite acceleration (to accelerate the object in an infinitely small time) but has a finite area (the area under the acceleration function is velocity). Sounds like the Dirac delta function, huh? In this scenario, the force applied to the object could be modeled as 2*m*delta(x), where m is the mass of the accelerated object. Physicists and engineers make the assumption that some things happen instantly, because they are so fast that trying to model them using actually equations would over complicate the problem, with no gain. So they use the Dirac delta function to make these "instantaneous" models. It may also help to think of the Dirac delta function as the derivative of the step function. The Dirac delta function usually occurs as the derivative of the step function in physics. In the above example I gave, and also in the video, the velocity could be modeled as a step function.
• I was wondering if there's a typo in the video, or if I'm misunderstanding something... if we say function equals 2*tau, while tau varies between - and +, doesn't it mean that the function graph goes under X axis (as the function value is negative when tau is negative)?
• Here, he's kind of assuming that Tau will always be positive (for the purpose of this video). Tau doesn't vary between - and + here, because of that implied assumption. Also, Tau isn't the independent variable here. The function is dependent on the variable t, or time. Because of that, Tau can be whatever Sal says it is no matter what t is.
• so does this mean that infinity * 0 = 1 ?
(1 vote)
• No, by definition ∞·0 is undefined. The Dirac delta function is a way to "get around" that, by creating a function that is 0 everywhere except at the origin, but the integral over the origin will be 1.
• So from to , you talk about shifting the delta function over the t axis.
But what happens if you, say, multiply t inside the delta function by a number? e.g. int (\delta (5t-3))? So in this example, I shift 3 units to the right, but then, when I evaluate integrals based on the definition that int (\delta(t) dt) = 1, I should divide by 1/5, right?

Also, what happens when we are talking about volume integrals on a delta function with a similar situation like the one postulated above? So int(\delta^3(5t-3))? I am not sure how I should "take the 5 out" to easily solve the integral. Should I also cube what I divide? ([1/5]^3)
(1 vote)
• The integral of delta(x) is = 1 IF the variable of integration is x, i.e.,
Int delta(x) dx = 1. Similarly, Int delta(x - 2) dx = Int delta(x - 2) d(x - 2) = 1.
But Int delta(2x) dx = Int delta(2x) d(2x/2) = (1/2) Int delta(2x) d(2x) = (1/2) 1 = (1/2). Similarly, for a > 0,
Int delta(ax + b) dx = (1/a) Int delta(ax + b) d(ax + b) = (1/a),
while for a < 0 we get,
Int delta(ax + b) dx = - (1/a) because the negative nature of a inverts the order of the integration.
(1 vote)
• Can you use he dirac delta function to model the collapse of a wave function for a particle? Please help
• "...he [Dirac] proposed the Dirac equation as a relativistic equation of motion for the wave function of the electron." Bibcode:1928RSPSA.117..610D.
(1 vote)
• is the same δ(t+1) = δ(1-t)
(1 vote)
• No, that is not true, \delta(1+t) has a singularity at -1 and \delta(1-t) has singularity at 1. I.e.
f(1)=\integral f(t)\delta(1-t)dt and
f(-1)=\integral f(t)\delta(1+t)dt
• Sal discussed the integral of the dirac delta function by giving a example of dtau(t) at , which fully explained the answer, but what if Dtau(t) isn't 1/2Pi at tau<t<tau, does it produce the same results?
(1 vote)
• Where were you in the '80s when I was forced to take this stuff. Those profs were as clear as mud - you make it look simple.
(1 vote)
• We didn't have internet, computers were pretty new, and professors were (relatively) rare. Not to mention, few people actually understood high level stuff like this. Even now, this is not common knowledge (not everyone majors in science), but this should be (perhaps even taught in high school?)
(1 vote)
• You shouldn't be able to use the regular Riemann integral on this particular function because the dirac delta function takes on certain values at discrete x values, so under these circumstances shouldn't we use a lebesgue integral? Since the number of all x's such that f(x) = infinity is countable (in this case 1 since the only x that makes f(x) = infinity is 0), that means its measure is 0, therefore making there area under which x makes f(x) = infinity 0. All the other x values make the function 0, so therefore it makes the area under the entire graph 0. Am I missing something?
(1 vote)
• You are missing that we can make the infinity as large as we want in order to make the integral of f(x) be 1. Thus, when using a Lebesgue integral, we would get A{0} + A{0} + A{0} + .... Since {0} is countable, its area is infinitesimal, but we also add it infinitely many times. To make the integral finite, we are allowed to say that we add the A{0} uncountably infinitely many times.
(1 vote)