The Dirac Delta 'Function': How to model an impulse or infinite spike

imagine i'm studying some system and then i just come along and hit it with a hammer what i've done is applied some force but i've applied it in a tiny tiny interval it's like being struck by lightning an enormous amount of energy but again a very tiny interval so as we study differential equations as we study the evolution of a system and time then we want to come up with a way to deal with these forces that are applied over very very tiny time intervals the way we're going to model this is by something called a delta function i want to begin with something called delta sub a comma epsilon which is a bit of a weird terminology but what i mean by this is well it has a graph of something that looks like this as in it's zero everywhere except in some interval a to a plus epsilon and then there the function is just a constant function that has height one over epsilon here i'm imagining epsilon is just some small number now if i do this i can then integrate the specific function and because it's just a rectangle the base is epsilon the height is 1 over epsilon they multiply out to give an area of one and since intervals are just the area under a curve the particular interval of this function is just one now i want you to imagine that i let that epsilon change so for example if i shrink it down a little bit i get something that is narrower and taller or i could go even more extreme and put the epsilon even a smaller value when i get a larger but thinner spike note that when epsilon is very small one over epsilon becomes very very large so this is indeed modeling the kind of hammer strikes that i'm imagining so now i want to take this to an idealized place i want to imagine epsilon going to zero and taking some sort of limit and so instead of it being zero most of the time and then just one over epsilon which is a big number i'm going to imagine instead that it actually goes to infinity at the value of a and is zero everywhere else because the integral of those delta functions when i was considering the epsilon were all one then i'm also going to have this being true as well that the integral from zero to infinity of this delta sub a is equal to one as well now what is this object it's not really a function because well the function value at a is not some number which is necessary to be a function but we can call it a generalized function or sometimes this is called a distribution if one wants to think of it for more sort of a probabilistic perspective and my goal here is not to try to delve too deeply into what this object really is but i just want to imagine it as something that's zero everywhere and then has an infinite spike specifically at the value of a and then additionally is going to have the property that when you integrate it out it's going to be equal to 1. in a sense we are going to define this so-called generalized function so that it has this nice interval property now before i continue further with this delta function i want to actually remind you of a theorem from firster calculus this is the mean value theorem not for derivatives as it's most often standard universal calculus but the mean value theorem for intervals the theorem goes as follows if you integrate some function over some interval then this is just equal to the height of the function f of c where c is some value in between a and b multiplied by this difference in the interval b minus a the way the theorem comes about is okay well let me just imagine a graph of some generic function between a and b then the integral of this curve from a to b represents well the area under the curve so there's a question which is how do i compute the area under the curve and visually i'm going to imagine that i have a rectangle instead and the point here is that the area of this rectangle is the same as the area of the original curve and specifically well the base of the area under the curve is b minus a the height of the rectangle is just going to be the value of f of c where c is some point in between a and b and indeed i just sort of eyeballed it but it looks to me like the area under the curve is approximately the same thing as the area of the pink sometimes the pink rectangle is above the curve sometimes it's beneath the curve those differences are going to so nevertheless the area of the rectangle is equal to the area under the curve aka the definite integral this seems quite reasonable to believe that it could be true and indeed it is always true provided your function f is nice and particular needs to be continuous on the interval after b okay so that was a fact back to delta functions i now want to study the integral of the delta function multiplied by some other function i'm going to do this in the same way by dealing with sort of finite approximations to the delta function delta a sub epsilon recall the graph as we saw earlier and as a result of that what it does is everywhere outside of a to a plus epsilon it just makes it equal to zero so i just get rid of that and the improper integral just turned into an integral from a to a plus e and then for the height of the delta function it was just defined to be 1 over e so i'm going to put that in let me give some space and get rid of the picture well then what is this by the mean value theorem for integrals that we just saw this is going to be the width of the interval a to a plus epsilon is just a width of epsilon multiplied by the integrand at some point so this is epsilon times one over epsilon times g of c and then what about that c well the c is some value in between a and a plus epsilon it's in the defined interval the epsilon's just cancel and so i'm just left with g of c so again c being some value between a and a plus epsilon well that's all fine but i want to actually now talk about the actual delta sub a function or generalized function if you prefer and so i'm going to take the limit as epsilon goes to 0. now there is no epsilon in this expression except for in the constraint on c recall that c had to be inside of the interval a to a plus epsilon and if the epsilon is going to zero that means that interval is getting smaller and smaller and smaller it's just saying that the c has to go to a in other words if i go all the way to the delta a function i will say that the integral of delta a multiplied by some other function g of t is just g of a so the way i think about this is that the delta function is the sort of infinite spike at the value of a but even though it's an infinite spike we had the normalization condition that the integral from 0 to infinity of this thing was nevertheless just equal to 1. so now if i take that and multiply it by some other function g of x what happens here is that at the spike it multiplies to one times well g of a and then everywhere else is going to be zero and so it just pulls out the value of the function just pulls out the g of a the final thing i want to talk about is the relationship between this delta or impulse function that we've been talking about and a function that we've seen previously in our series on laplace transform namely we've seen before step functions so the way the step function was defined was that it was zero to the left of a and then it jumped up to be the value of 1 at the value of a and was 1 from then on and indeed we've seen that this step function played an important role in our study of laplace transforms to solve differential equations now the step function is discontinuous so the derivative of the step function at the value of a doesn't really make sense it's not defined however if i imagine taking the derivative of the value of a what's really going on here is because it's going straight up it goes from zero all the way up to one it's kind of like having an infinite derivative and i know if this was back in first year calculus where we were actually talking about functions you'd say that the derivative of the step function at a did not exist you'd take some limit as h goes to zero and so it did not exist except we're talking about these sort of generalized functions here and so i'm going to not deduce a fact i'm going to define a fact am going to define the derivative of the step function to just be equal to the value of the delta function away from a this is perfectly fine the step function is just flat its derivative is zero as is the delta function that makes sense but at the problem spot of t equal to a this is just a definition it's a reasonable definition because as i say the step function goes straight up and so it's kind of like having an infinite derivative and the delta function is infinite at the value of a so some sort of consistency but nevertheless this is just a definition all right so that was the delta function or the impulse function we've now studied a few of its properties in the next video i actually want to use it to solve differential equations and so in particular we're going to figure out what is the laplace transform of this particular delta function and then we will use that laplace transform to help solve differential equations that involve a sort of an impact like hitting the system with a hammer

Channel: Dr. Trefor Bazett

Views: 11,012

Rating: 4.9825706 out of 5

Keywords: Math, Solution, Example, Dirac Delta Function, Laplace Transform, Differential Equations, Step Function, derivative, functional

Id: SxNVcCVj-3c

Length: 9min 25sec (565 seconds)

Published: Sat May 09 2020