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## Differential equations

### Course: Differential equations > Unit 3

Lesson 2: Properties of the Laplace transform- Laplace as linear operator and Laplace of derivatives
- Laplace transform of cos t and polynomials
- "Shifting" transform by multiplying function by exponential
- Laplace transform of t: L{t}
- Laplace transform of t^n: L{t^n}
- Laplace transform of the unit step function
- Inverse Laplace examples
- Dirac delta function
- Laplace transform of the dirac delta function

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# Dirac delta function

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

## Want to join the conversation?

- At7:49, I guess you could say he's building a "tau-er". Haha.(72 votes)
- Hi everyone, i wanted to know, why at16:39Sal modelled the force applied with a Dirac's Delta ? Can't understand the motivation, and why this is possible...

Regards(7 votes)- 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.(22 votes)

- 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)?(3 votes)
- 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.(3 votes)

- 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.(5 votes)

- So from10:50to12:20, 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(2 votes)
- "...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(2 votes)

- Sal discussed the integral of the dirac delta function by giving a example of dtau(t) at3:57, 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)

## Video transcript

When I introduced you to the
unit step function, I said, you know, this type of function,
it's more exotic and a little unusual relative to
what you've seen in just a traditional Calculus course,
what you've seen in maybe your Algebra courses. But the reason why this was
introduced is because a lot of physical systems kind
of behave this way. That all of a sudden nothing
happens for a long period of time and then bam! Something happens. And you go like that. And it doesn't happen exactly
like this, but it can be approximated by the unit
step function. Similarly, sometimes you have
nothing happening for a long period of time. Nothing happens for
a long period of time, and then whack! Something hits you really hard
and then goes away, and then nothing happens for a very
long period of time. And you'll learn this in the
future, you can kind of view this is an impulse. And we'll talk about
unit impulse functions and all of that. So wouldn't it be neat if we had
some type of function that could model this type
of behavior? And in our ideal function,
what would happen is that nothing happens until we get
to some point and then bam! It would get infinitely strong,
but maybe it has a finite area. And then it would go back to
zero and then go like that. So it'd be infinitely high right
at 0 right there, and then it continues there. And let's say that the area
under this, it becomes very-- to call this a function is
actually kind of pushing it, and this is beyond the math of
this video, but we'll call it a function in this video. But you say, well, what good
is this function for? How can you even
manipulate it? And I'm going to make one more
definition of this function. Let's say we call this function
represented by the delta, and that's what we do
represent this function by. It's called the Dirac
delta function. And we'll just informally say,
look, when it's in infinity, it pops up to infinity
when x equal to 0. And it's zero everywhere else
when x is not equal to 0. And you say, how do
I deal with that? How do I take the integral
of that? And to help you with that, I'm
going to make a definition. I'm going to tell you what
the integral of this is. This is part of the definition
of the function. I'm going to tell you that if I
were to take the integral of this function from minus
infinity to infinity, so essentially over the entire real
number line, if I take the integral of this function,
I'm defining it to be equal to 1. I'm defining this. Now, you might say, Sal, you
didn't prove it to me. No, I'm defining it. I'm telling you that this delta
of x is a function such that its integral is 1. So it has this infinitely
narrow base that goes infinitely high, and the area
under this-- I'm telling you-- is of area 1. And you're like, hey, Sal,
that's a crazy function. I want a little bit better
understanding of how someone can construct a function
like this. So let's see if we can satisfy
that a little bit more. But then once that's satisfied,
then we're going to start taking the Laplace
transform of this, and then we'll start manipulating
it and whatnot. Let's see, let me complete
this delta right here. Let's say that I constructed
another function. Let's call it d sub tau And this
is all just to satisfy this craving for maybe a better
intuition for how this Dirac delta function
can be constructed. And let's say my d sub tau of--
well, let me put it as a function of t because everything
we're doing in the Laplace transform world,
everything's been a function of t. So let's say that it equals 1
over 2 tau, and you'll see why I'm picking these numbers
the way I am. 1 over 2 tau when t is
less then tau and greater than minus tau. And let's say it's 0
everywhere else. So this type of equation,
this is more reasonable. This will actually look like
a combination of unit step functions, and we can actually
define it as a combination of unit step functions. So if I draw, that's
my x-axis. And then if I put my
y-axis right here. That's my y-axis. Sorry, this is a t-axis. I have to get out
of that habit. This is the t-axis, and, I mean,
we could call it the y-axis or the f of t-axis, or
whatever we want to call it. That's the dependent variable. So what's going to
happen here? It's going to be zero everywhere
until we get to minus t, and then at minus
t, we're going to jump up to some level. Just let me put that
point here. So this is minus tau, and
this is plus tau. So it's going to be zero
everywhere, and then at minus tau, we jump to this level, and
then we stay constant at that level until we
get to plus tau. And that level, I'm saying
is 1 over 2 tau. So this point right here on the
dependent axis, this is 1 over 2 tau. So why did I construct this
function this way? Well, let's think about it. What happens if I take
the integral? Let me write a nicer
integral sign. If I took the integral from
minus infinity to infinity of d sub tau of t dt, what is this
going to be equal to? Well, if the integral is just
the area under this curve, this is a pretty straightforward thing to calculate. You just look at this, and you
say, well, first of all, it's zero everywhere else. It's zero everywhere else, and
it's only the area right here. I mean, I could rewrite this
integral as the integral from minus tau to tau-- and we don't
care if infinity and minus infinity or positive
infinity, because there's no area under any of those
points-- of 1 over 2 tau d tau. Sorry, 1 over 2 tau dt. So we could write it this
way too, right? Because we can just take the
boundaries from here to here, because we get nothing whether t
goes to positive infinity or minus infinity. And then over that boundary, the
function is a constant, 1 over 2 tau, so we could just
take this integral. And either way we evaluate it. We don't even have to know
calculus to know what this integral's going
to evaluate to. This is just the area under
this, which is just the base. What's the base? The base is 2 tau. You have one tau here and
then another tau there. So it's equal to 2 tau
times your height. And your height, I just
said, is 1 over 2 tau. So your area for this function,
or for this integral, is going to be 1. You could evaluate this. You could get this is going to
be equal to-- you take the antiderivative of 1 over 2 tau,
you get-- I'll do this just to satiate your curiosity--
t over 2 tau, and you have to evaluate this
from minus tau to tau. And when you would put tau in
there, you get tau over 2 tau, and then minus minus tau over
2 tau, and then you get tau plus tau over 2 tau, that's
2 tau over 2 tau, which is equal to 1. Maybe I'm beating
a dead horse. I think you're satisfied that
the area under this is going to be 1, regardless
of what tau was. I kept this abstract. Now, if I take smaller and
smaller values of tau, what's going to happen? If my new tau is going to be
here, let's say my new tau is going to be there, I'm just
going to pick up my new tau there, then my 1 over
2 tau, the tau is now a smaller number. So when it's in the denominator,
my 1 over 2 tau is going to be something
like this, right? I mean, I'm just saying, if I
pick smaller and smaller taus. So then if I pick an even
smaller tau than that, then my height is going to be
have to be higher. My 1 over 2 tau is going
to have to even be higher than that. And so I think you see where
I'm going with this. What happens as the limit
as tau approaches zero? So what is the limit as tau
approaches zero of my little d sub tau function? What's the limit of this? Well, these things are going
to go infinitely close to zero, but this is the limit. They're never going to
be quite at zero. And your height here is going to
go infinitely high, but the whole time, I said no matter
what my tau is, because it was defined very arbitrarily,
was my area is always going to be 1. So you're going to end up with
your Dirac delta function. Let me write it now. I was going to write
an x again. Your Dirac delta function is a
function of t, and because of this, if you ask what's the
limit as tau approaches zero of the integral from minus
infinity to infinity of d sub tau of t dt, well, this should
still be 1, right? Because this thing right here,
this evaluates to 1. So as you take the limit as tau
approaches zero-- and I'm being very generous with
my definitions of limits and whatnot. I'm not being very rigorous. But I think you can kind of
understand the intuition of where I'm going. This is going to
be equal to 1. And so by the same intuitive
argument, you could say that the limit from minus infinity to
infinity of our Dirac delta function of t dt is also
going to be 1. And likewise, the Dirac delta
function-- I mean, this thing pops up to infinity at
t is equal to 0. This thing, if I were to draw my
x-axis like that, and then right at t equals 0, my
Dirac delta function pops up like that. And you normally draw
it like that. And you normally draw it so
it goes up to 1 to kind of depict its area. But you actually put an arrow
there, and so this is your Dirac delta function. But what happens if you
want to shift it? How would I represent my--
let's say I want to do t minus 3? What would the graph
of this be? Well, this would just be
shifting it to the right by 3. For example, when t equals 3,
this will become the Dirac delta of 0. So this graph will just
look like this. This will be my x-axis. And let's say that this
is my y-axis. Let me just make that 1. And let me just draw some points
here, so it's 1, 2, 3 That's t is equal to 3. Did I say that was the x-axis? That's my t-axis. This is t equal to 3. And what I'm going to do here is
the Dirac delta function is going to be zero everywhere. But then right at 3, it
goes infinitely high. And obviously, we don't have
enough paper to draw an infinitely high spike
right there. So what we do is we
draw an arrow. We draw an arrow there. And the arrow, we usually draw
the magnitude of the area under that spike. So we do it like this. And let me be clear. This is not saying that the
function just goes to 1 and then spikes back down. This tells me that the
area under the function is equal to 1. This spike would have to be
infinitely high to have any area, considering it has an
infinitely small base, so the area under this impulse function
or under this Dirac delta function. Now, this one right here is t
minus 3, but your area under this is still going to be 1. And that's why I made
the arrow go to 1. Let's say I wanted to graph--
let me do it in another color. Let's say I wanted to graph
2 times the Dirac delta of t minus 2. How would I graph this? Well, I would go to t minus 2. When t is equal to 2, you get
the Dirac delta of zero, so that's where you would
have your spike. And we're multiplying it by 2,
so you would do a spike twice as high like this. Now, both of these go to
infinity, but this goes twice as high to infinity. And I know this is all being
a little ridiculous now. But the idea here is that the
area under this curve should be twice the area under
this curve. And that's why we make the arrow
go to 2 to say that the area under this arrow is 2. The spike would have to
go infinitely high. So this is all a little
abstract, but this is a useful way to model things that are
kind of very jarring. Obviously, nothing actually
behaves like this, but there are a lot of phenomena in
physics or the real world that have this spiky behavior. Instead of trying to say,
oh, what does that spike exactly look like? We say, hey, that's a Dirac
delta function. And we'll dictate its impulse
by something like this. And just to give you a little
bit of motivation behind this, and I was going to go here in
the last video, but then I kind of decided not to. But I'm just going to show it,
because I've been doing a lot of differential equations and
I've been giving you no motivation for how this applies
in the real world. But you can imagine, if I have
just a wall, and then I have a spring attached to some mass
right there, and let's say that this is a natural state
of the spring, so that the spring would want to be here,
so it's been stretched a distance y from its kind of
natural where it wants to go. And let's say I have some
external force right here. Let's say I have some external
force right here on the spring, and, of course, let's
say it's ice on ice. There's no friction
in all of this. And I just want to show you
that I can represent the behavior of this system with
the differential equation. And actually things like the
unit step functions, the Dirac delta function, actually start
to become useful in this type of environment. So we know that F is equal to
mass times acceleration. That's basic physics
right there. Now, what are all
of the forces on this mass right here? Well, you have this
force right here. And I'll say this is a positive
rightward direction, so it's that force, and then
you have a minus force from the spring. The force from the spring
is Hooke's Law. It's proportional to how far
it's been stretched from its kind of natural point, so its
force in that direction is going to be ky, or you could
call it minus ky, because it's going in the opposite direction
of what we've already said is a positive
direction. So the net forces on this is F
minus ky, and that's equal to the mass of our object times
its acceleration. Now, what's its acceleration? If its position is y, so if y
is equal to position, if we take the derivative of y with
respect to t, y prime, which we could also say dy dt, this
is going to be its velocity. And then if we take the
derivative of that, y prime prime, which is equal to d
squared y with respect to dt squared, this is equal
to acceleration. So instead of writing a, we
could right y prime prime. And so, if we just put this
on the other side of the equation, what do we get? We get the force-- this force,
not just this force; this was just F equals ma-- but this
force is equal to the mass of our object, times the
acceleration of the object plus whatever the spring
constant is for the spring plus k times our position,
times y. So if you had no outside force,
if this was zero you'd have a homogeneous differential
equation. And in that case, the spring
would just start moving on its own. But now this F, all of a
sudden, it's kind of a non-homogeneous term, it's what
the outside force you're applying to this mass. So if this outside force was
some type of Dirac delta function-- so let's say it's t
minus 2 is equal to our mass times y prime prime plus our
spring constant times y, this is saying that at time is equal
to 2 seconds, we're just going to jar this thing
to the right. And it's going to have an-- and
I'll talk more about it-- it's going to have
an impulse of 2. It's force times time is going
to be-- or its impulse is going to have 1. And I don't want to get too much
into the physics here, but its impulse or its change in
momentum, is going to be of magnitude 1, depending on
what our units are. But anyway, I just wanted to
take a slight diversion, because you might think Sal is
introducing me to these weird, exotic functions. What are they ever going
to be good for? But this is good for the idea
that sometimes you just jar this thing by some magnitude
and then let go. And you do it kind of infinitely
fast, but you do it enough to change the momentum of
this in a well-defined way. Anyway, in the next video, we'll
continue with the Dirac delta function. We'll figure out its Laplace
transform and see what it does to the Laplace transforms
of other functions.