hi again everyone Chris Tisdell here again in this presentation I'm going to continue my series of videos on partial differential equations in particular I'm going to show you how you can solve some problems from PDE using Laplace transforms now in other videos I've shown you how to solve our ot YZ and associated initial value problems via an application of Laplace transforms and so this naturally raises the question whether you can solve PDE s via this this method and the answer is yes so this is the subject of today's video and we're going to solve an example which is a linear PDE it's just going to be first-order and some associated boundary conditions okay so this is the problem that we're going to look at now you can see here that one of the claims I've made is that as the Laplace transform techniques are well suited to linear PD with constant coefficients well here the coefficients not constant but it's a function of X and not T and because we transform both sides with respect to T then you can treat X as a constant so coefficients and X are okay now just to refresh your memory or if you haven't seen this this method before up let me give you an overview of what happens you start with a PDE and some conditions you transform the problem via the Laplace transform you have denoted it by this curly L and what you do is you produce a transformed problem usually an eau de now what you do is you solve this transform problem where this W bar represents the transform of the solution W to the original problem and then what you do is you transform everything back to recover the solution to the original problem okay so this to me is a general guide for solving problems via a Laplace transform methods or just transport methods in general okay so we'll get to this problem a little bit later but the first question is what is the Laplace transform well it's an integral transform if you haven't seen the plastron forms before I just wanted to give you a quick introduction and for those of you who have just a quick reminder so the suppose I've got a function of one variable so this might T might represent time and this might be a solution to a certain ordinary differential equation what what you do is you take the transform you take the original function f of T multiply through by an exponential and you integrate with respect to T over the half line now with a regards to notation lowercase letters are usually associated with the original function of T and uppercase or capital case letters are associated with the transform with a lowercase letter function now you can see in here the we integrate with respect to T but there's an S in there as well so this notation emphasizes that actually if we for example integrated this you would be left with a function of s ok so what happens if we have a function of two variables so this W of X comma T might be a solution to a partial differential equation well you can still define the Laplace transform very similar except you have a function of two variables in here and again we use capital edits now just to be very short I'm also going to put a bar over the double here because sometimes when I write W lowercase W and uppercase W it looks very similar especially when I'm writing it out by hand so there's no confusion I'm going to use a bar over the capital letter to denote the transform of the lowercase letter function okay now if you want to you can you know play around with this and put in various functions of T and compute this integral or what's what's also very common is using a table to calculate transforms now something I haven't mentioned yet the f of t satisfies its piecewise continuous and it satisfies some sort of exponential growth condition so the Thea the improper integral will converge okay even though I haven't listed that all right some important properties for out investigation today is going to be the transform of derivatives so in all of these transforms I'm transforming with respect to the T variable okay so you can prove these two expressions just using integration by parts so you would put up here you would replace W with W sub T where the sub T denotes partial differentiation with respect to T and then you'll just integrate by parts okay so in this sort of Big W bar notation this would be OOP sorry that's it just s this would be the following and down here this would be the following so says what happens if we take the Laplace transform a second-order derivative with respect to T and what happens if we're taking the Laplace transform with respect to T of an X derivative well it's just the derivative of the transform with respect to T ok so to prove that you can use Leibniz rule for differentiating under the integral sign so if I was to take this and so I just got there and replace this with w sub X see if I can squeeze it in I'm just going to drop off the argument that should save a bit of space now this derivative can be shifted outside the front of the integral sign so I'm sort of using Leibniz rule in Reverse here okay what is this this is just this so it's the derivative of the transform okay and you know if you're transforming the second derivative with respect to X then that would just you never d squared DX squared okay so these are important properties known as transforms of derivatives that we'll use and apply when we're solving our problems viola plas transforms okay so let's have a look at the following problem solve a Laplace transforms okay so we've got a first-order problem here coefficient is a function of X you've got some extra conditions known as boundary conditions and let's just follow this model here transform the PDE produce a transform problem which is essentially an Oda an ordinary differential equation solve for the solve the transform problem and then invert everything by on inverse transform okay now just to mention the inverse transform is an operation that undoes little past transform so if I transform something and take the inverse transform I get what I started with in and you can switch it around okay now the inverse transform to really write down a an explicit representation for it you need an understanding of complex analysis so I'm going to assume that you don't have a deep understanding of complex analysis so we can do these problems without knowing exactly what the inverse transform is okay all right so let's follow this guideline here let's transfer everything and produce some sort of transform problem all right so I transform both sides and because it's an integral operator and it sends its a linear linear operator I can break up the left-hand side and I even though I'm taking the transform of this we're transforming with respect to T so we imagine that X's is essentially a constant here transform of zero is zero now what I can do is simplify this a little bit by using my transform of derivatives okay so up here I'll get that and I've got this as well so Tod is going to stand for transform of derivatives okay so over here this is just the derivative the derivative of the transform and this here is this minus this again I've just left off the arguments here and here for to save a bit of space okay let's look up to our conditions up here we can use this in here we know that this is 0 so that term is going to disappear since this can be zero and we get the following so we can see that we've actually reached this stage now we've produced a transform problem which is essentially an Eau de why is it an Eau de because it's a PDE where only one derivative is present okay so let's see if we can solve this transform problem so that's what we're trying to do now here at this stage so let me just make a comment there this is essentially and Oh de it's first order linear and separable so what we can do is solve this just using ordinary differential equations techniques so how do we do it well we know that the solution to this is going to be something like an exponential function okay so imagine s is a constant here okay we want to solve this for W bar okay so you look at the coefficient of W bar and because it's on that on the left hand side of the equation you would take e to the integral of the negative of that coefficient okay and then you would have what's something like a constant of integration here but because W bar is a function of two variables this really should be a function of s okay so if I do this integral with respect to X I'll get the following I can imagine eight is a constant all right so I now have a general solution to this problem so I've almost solved for W behave almost solve the transform problem but I haven't found this a of s now I've used this I've used this but I haven't used this yet so to find this this coefficient we're going to use this okay so let me just this is the general solution w bar okay now let's obtain a of s by the following well if I transform this I'll get W bar of zero comma s okay what is the Laplace transform of T well it's one on s squared now you can work that out either directly or from a Laplace transform table okay so now what I can do is go up here and go okay well when x equals zero in here I'm going to have e to the 0 times a of s which is just a this and I know from this condition that W bar of 0s is one of the squid so I could team those up okay so that and that with x equals zero gives us my coefficient okay so we've now solved our transform problem so we've completed this step let's move to recover W the solution to the original problem we've solved the transform problem let's get back to the original setting now ok so to do this we need to take an inverse transform ok so how do we do that well we want to take the inverse transform of both sides now there is a theorem from the from Laplace transforms which gives you the inverse transform of a function of x times an exponential function ok that is called the second shifting theorem ok so we're going to apply that and we'll get our solution W okay by the second shifting theorem okay what is the second shifting theorem it's the following if I have function of s times an exponential I want it here a is a constant and I want to take the inverse transform it's the following okay the inverse transform of this product is just a Heaviside step function shifted times G of T shifted where G of T is the inverse transform of this function okay so for this G of s would be one on s squared a would be x squared and essentially what you want to do is compute the inverse transform of one on s squared shift it and then multiply through by this Heaviside step function okay so let me show you how to do that oops okay so G of T equals the inverse transform one on s squared now the inverse transform of one of s but s squared is T you can look that up from a table and so applying the second shifting theorem do you transport this it's this shifted so I'll replace a with x squared and then times G of t minus a so G is G of T is T so I replace T with t minus x squared okay so what is this this unit step function or Heaviside step function well when t minus x squared is less than zero this thing will be zero so the whole thing will be zero when t minus x squared is greater than 0 this becomes 1 so the whole thing would be sorry G is just the G GG is just the identity function here ok so you don't need that G there okay let me put in some square brackets okay so this will be 0 when t minus x squared is less than 0 it'll be 1 times this when t minus x squared is greater than 0 and what happens at when t minus x squared equals 0 well in my videos I define this to be 1/2 but it doesn't really matter because that's going to be 0 anyway so you get 1/2 times 0 so it's just 0 okay so I've managed to squeeze everything onto that page the you can see now that we are at the final stage we've recovered W the solution to our original problem using an inverse transform ok so there you have it now the steps for these kinds of problems are the same start with your PDE transform it include some of the conditions produce a transform problem which you can solve by incorporating other conditions once you've got your solution to the transform problem you transform everything back by an inverse transform okay and you recover the solution to the original problem so going from here to here tables are useful ok sometimes you can compute inverse transforms without tables for example using residues from complex analysis but in this presentation I'm just assuming that you're that you're working from a table ok so I hope this presentations been useful in upcoming videos I'll solve some second-order problems now I know you can solve this this problem perhaps using method of characteristics or other other techniques so what I'm going to do is show you how to apply the Plast Rance form to solve second-order problems I hope you can join me for those presentations