Intro to the Laplace Transform & Three Examples

welcome to my new series on the Laplace transform in this video with videos going forward we're going to study some of the reasons why the Laplace transform is an incredible powerful piece of mathematics in this first video were to define little past transform only see a couple different examples where we're going to compute the Laplace transform but why wouldn't we be interested in something a little past transform in the first point what reason why I really like the Laplace transform is it's useful for solving differential equations like this one here I have an ordinary differential equation together with two different initial conditions in general differential equations can be very challenging although this specific one can be solved as is a constant coefficient homogeneous one relatively straightforward whatever the point of the Laplace transform is to convert differential equations like this ones that have two rivers in them into somehow an algebraic equation this is some new equation it's got a new Y in terms of a new variable s but there's no tourism fault it's just a purely algebraic equation in d in this particular case is very simple to solve for y I can just divide through and so one of the powers of the Laplace transform is going to be able to convert differential equations into algebraic equations that were able to solve them then we're going to learn how we can convert backwards and take the solution to the algebraic equation and figure out a solution to the original differential equation so that's our plan for the future but right now in this first video we're simply going to define the Laplace transform and we're going to compute what the Laplace transform is in three different examples the notation for the Laplace transform is I through the sort of squiggly owl and what this squiggly L does is it has an input which is some function of T and then you're going to take that F of T and via the Laplace transform you're going to transform it into some other function now a capital F and this other function depends on some other variable now s so the last transform is is kind of like a function which would take a to a point but the Laplace transform takes a function to a different type of function here's how is defined the Laplace transform this capital F of s is defined to be the improper interval from 0 to infinity of e to the minus st notice that's where the S comes in of F of T DT because we're doing an interval with respect to T the T is going to go away so that what you get for your f of s is it only depends on the value of the ass this thing that comes in by this negative exponential because this is an improper interval we're going to have questions like when does this converge and when visit not converge nevertheless this is the definition so let's see some examples the first one I want to begin with is just going to be the Laplace transform of an exponential function so I have e to the a T where a is just some constant all right so what's its Laplace transform we can just simply write it down so this is the improper integral from 0 to infinity e to the minus st I just copy that and then for the F of T I plug in and I plug in the e to the a T this is two different Exponential's so I can combine the Exponential's the powers of them and make it just a minus s times T this is a straightforward enough integrand so when you integrate that you're going to get e to the a minus s times T and then have the divide out by a minus s and we're evaluating this between 0 and infinity by the way because we're doing improper integrals the technical way to define these is to replace the infinity sign with say a B and then people limit as B goes to infinity nevertheless we have this shorthand to sort of evaluate at 0 and infinity ok so putting this together and getting rid of the mess this is my claim how do I evaluate that well it depends on the a and the s the question of whether this is going to converge when you take this limit as some value goes to infinity is that going to be a finite number or not well it depends on the a and the S so for example if s is greater than a we're going to get one case in that is less than or equal to a look at another when s is greater than a in the exponent you have e to well a negative number times T as T goes to Entebbe his rookie just a really windy and howling upside but you know when we're in the middle of go good night D I got a record the video down nevertheless so you can hear some rendered wind noises just ignore that anyways and then if you take T to infinity of a negative exponential it just goes away to zero and so indeed in this case it just converges as zero and then when you plug in this early good and e to the zero over the a minus s and and the end just 1 over s minus n in the other case where s is less than or equal to 8 well if it's equal then just in the denominator you have division by zero diverges and if the S is strictly less than 8 it's going to be well it's going to be an exponential that has a positive times at T going to infinity likewise it's going to diverge so in this example we figured I'd help to compute the Laplace transform of this exponential function we've noted that the answer that we get does depend on what the value of s is relative to this constant a and you sometimes get it divergent and sometimes get it converging before we get to our second example I want to introduce a new function to us I'm going to call this a step function or sometimes called the Heaviside function what this function does is it's 0 when T is less than 0 and it's 1 when T is bigger than 0 and then the example I want to compute out is actually the Plast rands form not just at the step function but of the step function U of T minus a but this U of T minus a looks like as well something like this the idea is that when your T is going to be less than a that is just going down to 0 and then when your T is greater than a it's going to the height 1 this is very useful function because if you have any function that you think should have a discontinuity at some particular spot when you multiply by this U of T minus a where a is the spot where you want to have a discontinuity it introduces a disk and so this is a wonderful way to be able to model a lot of functions that have step discontinuities you just sort of multiply by this heavy side function or this step function nevertheless our goal is to compute the Laplace transform of this function well let's write it down so again it's our integral but now our integral doesn't start at zero to infinity because the function is just the function zero all the way up to the value of a so the portion of the integral that's zero up to a is just going to be equal to zero here by the way I'm assuming that a is a positive constant so the integrand has that negative exponential that is part of the definition of the Laplace transform and then it just multiplies by one but what really is happening with the step function is it's restricting the domain so it's now if a to infinity not 0 to infinity nevertheless it's an easy integrand we can compute that and when you plug in the Infinity because it's a negative exponential it goes away to zero our assumption here is that a is positive and as a result it's just going to be e to the minus s a divided about a it's what happens when you plug in the lower limit that T is equal to a alright so now let's do our third example but before we do that I want to introduce a new piece of terminology I want introduce something called the gamma function the gamma function looks a little bit relegated to the Laplace transform indeed it will be in just one moment but for now it's just an improper interval from 0 to infinity very similar the specific exponential e to the minus T and then multiplied by T to the X minus 1 DT this is some function and it turns out to be important enough will be give it a name and so we call it the gamma function ok the give a function has actually many very pleasing properties the first sort of simple one is just that the gamma function of 1 turns out just to be 1 if you plug in X equal to 1 then you get T just to the 0 we just comes a negative exponential you can integrate that and get 1 but perhaps more interestingly is and what you will see what happens to the gamma function when I evaluate it at an integer n plus 1 so if I do that and I plug this in well T to the n plus 1 minus 1 it becomes a 2 so now this is some sort of expression here if you thought about how you might have to integrate this well you could do integration by parts and different times to produce that T to the N down to a 1 but I'm actually only gonna do it once and try to get some sort of recursive behavior that I can use so that's my going to do I'm gonna set you legal T to the end I'm gonna set DV equal to e to the minus T DT and I'm gonna do it in two richmond parts what this gives me is the following long expression so I get au times of e evaluated the end points throw to the infinity and then I subtract off a V to you and that gives me this expression now for the first part that's evaluated at infinity and 0 well e to the negative infinity is going to dominate the polynomial and is going to give a 0 and when you plug in 0 you get well just zero to the end you're like was gonna get circle so the the first portion of this is just all 0 what about the second portion well the second portion kind of looks like another gamma function there isn't any mixed in there so we have to take that in and bring it out but other than that is just n times the gamma function of M so what I've done here is I have related the gamma function of n plus 1 to the gamma function of n and it brought out this coefficient and I could do the exact same argument again and I can say well look I'm gonna take the gamma and related to gamma of n minus 1 and that will bring out an N minus 1 and then I can just keep going in this way time and time again going all the way down gamma and gamma n minus 1 gamma n minus 2 and so forth all the way down to 1 we've seen the gamma 1 with 1 and so really what we just have it just n factorial and n minus 1 n minus 2 and so forth so it's quite pleasing that for n factorial which has ever made a function that we've thought of is sort of this thing you could go say it throws up or integrate now can be expressed as this particular improper integral one of the next things that we have is a bit of a generalization of the factorial for example you can ask what gamma of say 1/2 is you don't have an answer for what what 1/2 factorial should be defined as but now you can express it in terms of this particular interval you know that interval it turns out has some very nice properties alright so with the gamma function defined now let me turn to this problem of computing the Laplace transform of just a polynomial T to the N it turns out that we're going to be using the gamma function in our computation indeed if we write out what our expression is going to be equal f of s the Laplace transform it just be improper and go e to the minus s T times T to the N DT I'm gonna try to clean this up a little bit to make it look more like the gamma function by doing the substitution u equal to s times T if I do that well the exponential just become e to the minus u but then further I can say that the T to the N is going to become a you at the end but it divides out by an S to the end goes on the bottom of it and then let wise when the DT turns into a duu also divided by one more copy of s so what you get is 1 over s to the n plus 1 snuck out the front and then some integrals entirely in terms of U now we can recognize this integral as just being the gamma function so I have that same coefficient on the front but now I just have the gamma function evaluated at n plus 1 and we've computed that out that was just n factorial so our final answer for the Laplace transform of T to the N is n factorial over s to the n plus 1 so that is that for so those are my three examples of computing out Laplace transform and in the next video we're going to see three different properties and Laplace transform paths if you have questions about this video leave them down in the comments you will see more differential equations or any weather courses the links to those places will be down in the description give the movie polite or that you developer them we'll do some more math in the next video

Channel: Dr. Trefor Bazett

Views: 678,527

Keywords: Math, Solution, Example, Laplace, Transform, Equation, Step Function, Heaviside, Heavyside, Gamma Function, exponential, compute, Differential Equation, ODE

Id: KqokoYr_h1A

Length: 12min 5sec (725 seconds)

Published: Sun Mar 15 2020