 Today we are going to start the second chapter, it is called first order partial differential equations. In this chapter we will study first order partial differential equations and Cauchy problem associated to that and what are the possibilities in terms of existence, uniqueness and maybe non-existence as well we will discuss. Now in today's lecture what we are going to discuss is how the first order PDEs appear, how they arise and the Cauchy problem, initial value problem and initial boundary value problems, what are they we will discuss briefly. So the outline for today's lecture is the first point is how the first order PDEs appear, we will see two situations, one of them is going to be a physical situation and the other one is within mathematics, mathematical situation. And then we define these terms Cauchy problem, initial value problem, initial boundary value problem for first order PDEs, we look at 3 examples which suggest us what happens or what can happen to a Cauchy problem in terms of existence of solutions, uniqueness and then we look at a simple first order PDE for which we consider IBVP that is initial boundary value problem and we study what are the possibilities there. Now how the first order PDEs appear, so first order PDEs arise in many situations, we are going to just see two instances, the first one being as a mathematical model for a physical phenomena, in fact what we are going to see is not really a physical phenomena it just models the traffic on a highway and then we see as equation satisfied by some mathematical objects. So this is the physical model which we are going to look at today is a mathematical model for traffic flow. So in this model we are going to have two main ingredients, we will briefly introduce them. So one of the most fundamental physical principles is what are called conservation laws or balanced laws and one encounters them already in course of mechanics where we would have seen conservation of mass, energy and angle momentum. So we are going to derive a PDE model in a similar situation, we call it simpler because no a priori knowledge of anything is needed including physics. As I told you we are not going to do really a physical model, it is going to actually model of the traffic flow which can be easily understood. So let us consider an express highway without any intermediate entry or exit points. It has one entry point, one exit point, you cannot join the expressway in between or you can leave, both possibilities are not there, leaving and joining is not possible. So the traffic what we mean is let us say vehicles, let us say cars just to refer very easily, assume that vehicles are moving in a straight line. As we know vehicles do not move straight lines, they will be overtaking one another but what this means is that this gives us a way to identify the highway as an interval in R and it actually physically means that overtaking is not allowed, maybe the expressway is very narrow so you cannot overtake. Now let rho denote car density at a point x in R, yes, rho is not infinite but then we are going to finally specialize to certain intervals. So at the time instant t, what does this mean, car density means what? We can get the number of cars by integrating with respect to x. So let a be less than b, then the number of cars in the interval a b in terms of the highway it may be a stretch between a point a and a point b, at time t is given by the integral of rho over the interval a b. Now we need to know how it changes, so the instantaneous rate of change at time t is given by the derivative of this number d by dt of the integral a to b rho x t dx. But who is really responsible for this change, it is a cars which are entering at the point a and leaving from the point b. Therefore the same number is given by the influx of cars at a minus outflux of cars at b which is qat minus qbt. Both are same numbers because we made this assumption of no entry and no exit. Thus we have this balance law, so both these quantities are balanced therefore d by dt of integral a to b rho x t is equal to qat minus qbt that is the balance. So the balance law is now d by dt a to b rho x t dx equal to qat minus qbt. Now we are going to modify the right hand side, the right hand side we write it as an integral. So if you use this fundamental theorem of calculus will give you qat minus qbt. So interchanging the order of differentiation and integration in the first term d by dt is outside the integral we can take it inside. Yeah of course we know that this is not true all the time it is not allowed one needs justifications but this where I say no bargaining ends at the very first instance. So do not ask that questions right now, they can be asked later. So now once the d by dt go inside the integral the balance law becomes a to b rho by dou t equal to minus integral a to b dou q by dou x. Now let us take the RHS to the left hand side and then the balance law becomes integral a to b of this quantity dou rho by dou t plus dou q by dou x equal to 0. Now this is true for every a and b such that a less than b. In other words you have integral of certain quantity is 0 over every interval a b it means that integrand has to be 0. This is an exercise in analysis we would have encountered many such situations for example if you take a continuous function integral integrate on any interval a b and if the result is 0 then the continuous function has to be 0. And there are generalizations of this which may be let us say locally integrable function if you integrate on any set and the integral is 0 then the function must be 0. So these are two typical instances where we can conclude this. Now what remains is the equation has a row that is what is the traffic about we want to study about this but there is a unknown quantity q the flux it is not known therefore we need to model that. Therefore it is reasonable to think that q depends on x time t and the density of course itself that is rho. So q can be thought of as a function of x t and rho this kind of proposition of what kind of a function it is of the variables x t and rho is what is called a constitutive law. Of course there will be as many models as the number of constitutive laws. So constitutive law models how q depends on x t rho consider a simplest constitutive law which is given by q of x t rho since x t is not involved in this formula I have not written x t. So q of rho equal to c rho where c is a positive number. This means the flux is proportional to the number of cars. So substituting this in the balance law what we get is dou rho by dou t plus c times dou rho by dou x equal to 0. This equation is called a linear transport equation. This is a linear equation is clear because the unknown quantity rho itself does not appear what appears is their first order derivative dou rho by dou t and dou rho by dou x it appears with the power 1. So this is a first order linear equation c is a constant. Therefore it is a linear equation why is it called transport equation something must be transported from somewhere to somewhere it must be from some time t equal to something to future times. So in deriving this model recall that there were two main ingredients first one was balance law and second one is a constitutive law or constitutive relation. Balance laws are very general constitutive relations model the specificness of the situation in this case of the highway. So in our model we use the constitutive relation as q rho equal to c rho in general q may be more general function of x t and rho in that case the balance law becomes this dou rho by dou t plus dou by dou x of q of x t rho if we expand this you will get this one. So you need to differentiate x appears here as well as in rho therefore need to differentiate q with respect to the first variable and differentiate q with respect to the last variable and differentiate rho with respect to x this is a chain rule. So chain rule is a very important rule that one must be very comfortable when you want to do anything with differentiation. So this is a quasi linear equation it is a first order PDE quasi linear because this may depend on rho that is why it is a quasi linear equation. Now we are going to model a highway where entries and exits are there in between that means we will see now how the earlier model derived is going to change. So the entries and exits are also called source and sinks in the PDE literature. So let us denote the source sink density at x at the time instant t. Now the model becomes this is the rate of change earlier this was just equal to the flux but now this will also have the sourcing term which is this integral a to b fxt dx. Now bringing everything to one side what we get is integral a to b of certain thing is 0 and this is true once again as before for every n b such that a is less than b both are arbitrary therefore the integrand must be 0 which will be the case whenever the integrand is continuous or local integrable but it is not unreasonable to assume that integral of over any set is 0 implies the integrand is 0. Now let us look at the how the first order PDE is appear in a mathematical context. So we are going to see an example consider two parameter family of surfaces. In fact what we are going to consider is surfaces which are graphs z equal to f of x y a b what are a and b they are some parameters varying in some intervals. So it is a two parameter family of graphs of functions. Now we want to derive a differential equation satisfied by this family and we are saying partial differential equation therefore what we can only do here is to differentiate this with respect to the two independent variables x and y and this is the expression we get. Now assume that from these two relations you can solve a as a function phi of x y z x z y and the b as a function of psi x y z x z y. Now question is it always possible these are all questions but here if it is possible we will go ahead and do the next step. What is the next step I am going to substitute for a and b these formulas inside this then what I will have is z equal to f of x y phi of x y z x z y comma psi of x y z x z y which will be a nonlinear partial differential equation in general. Now let us look at a specific example where the functions taken to be x minus a whole square plus y minus b whole square a and b are parameters varying in real numbers both of them are real numbers. Now what are these surfaces what does this equation represent? Represents cones family of cones. So differentiate this relation with respect to x and y and eliminate a and b in other words solve for a and b in terms of rest of the things. So here it is very simple a equal to x minus z x by 2 b equal to y minus z y by 2 now substitute the values of a and b inside this expression z equal to x minus a whole square plus y minus b whole square which is the equation of the family of surfaces. Once you do that and simplify you get this following PDE z x actually stands for dou z by dou x if you want to write in terms of x you can write dou u by dou x whole square plus dou u by dou a whole square equal to 4 u. Now let us see what are the problems which goes by the name of Cauchy or initial value or initial boundary value problems what they are. So let us consider a PDE f of x y u x u u x u y equal to 0 it means I am considering a PDE in 2 independent variables throughout this chapter we are going to consider first order PDE in 2 independent variables and then we will also say what how theory changes if you have one more variable. So x y are independent variables u is the dependent variable these are the derivatives. So there are 5 quantities here so f must be defined on a subset of r 5 and that we denote as omega 5 is a subset of r 5 and you have a function and then this will then define a partial differential equation. So what is the Cauchy problem given a space curve gamma in r 3 that means you are given a curve in r 3 which is described parametrically by this that is x y z equal to f s g s and h s as s varies in some interval i and f g h are c 1 functions on the interval i and such that the projection gamma 2 of gamma to x y plane where is gamma it is an r 3 so it will have x y z coordinates. Now you project it to the x y plane you will have x y coordinates that is x equal to f s and y equal to g s that is the curve gamma 2 s in c m i it is a regular curve what that means is f dash g dash do not vanish that means does not become 0 0 at every point of the curve if one of them is 0 other one is non-zero if f dash is 0 g dash is non-zero and find a solution to the PDE such that u of f s g s equal to h s for s belonging to a sub interval of i sub interval of i in other words a part of gamma lies on this surface s denoted by s equation is set of all x y z such that z equal to u x y on this surface a part of this curve given curve gamma lies right gamma coordinates are f s g s and h s third coordinate h s must be equal to u of first 2 coordinates f s g s that is the condition here we are not requiring that this should happen for every s in i we are only asking for s belonging to sub interval of i so this can be thought of as a local solution. So this gamma is called a data curve or datum curve or initial curve note that this problem actually makes sense only if gamma is subset of omega 3 what is omega 3 is a projection of omega 5 which is in r 5 to the first 3 coordinates x y u what is initial value problem it is a special type of Cauchy problem initial means there is something like a time in the problem and initially means at some particular time in the past something is happening and then you are interested in studying the equation for future times. So therefore one of the variables will have an interpretation of time let it be y there is no loss of generality in assuming y x would equally do the same thing so the y variable has an interpretation of the time variable and the datum curve now lies in the zx plane given by this x equal to s y equal to 0 z equal to hs in other words what we will be asking is actually u of x, 0 equal to h of x should be satisfied this is what we are asking for the for the solution of the PDE. So find u of x t such that u t plus 2 u x equal to 0 and u x 0 equal to sin x parametric representation of gamma what is gamma x equal to s y equal to r here t equal to 0 it is not y it should be t because y has a interpretation of time variable so you can as well write t so t equal to 0 and z equal to sin s and s belongs to r that is a parametric representation of gamma. Now we do not know how to solve this for now but given a formula we can always check that it is a solution so u of x t equal to sin of x minus 2 t if you quickly differentiate you will get that u t plus 2 u x equal to 0 it solves the initial value problem. So when t equal to 0 this formula u of x 0 what will we get u of x 0 will be sin x so initial condition is satisfied so we can think of like this so this is the time variable this is the u variable this is the time is x at t equal to 0 you are given a sin x like that so you have to imagine this in the ux plane and then at t equal to half what you will have is a formula from the formula what we have is it is sin x minus 1 so u of x, half is sin x minus 1 it just means the graph of sin has been shifted to x equal to 1 suppose this is 1 this is pi right so this is pi by 2 so this is 1.5 already so somewhere here so the graph will look like this is moving. So it is difficult to write for me in 3 dimensions but you should imagine that the graph of sin x has moved graph was like this at t equal to 0 now it has shifted to 1 but now this I am writing at t equal to half so just moves back. So okay now initial boundary value problems once again initial is there so there should be a time so the unknown u in this PDE x, t, u, u, x, u, t equal to 0 I already made this change of y to t because the word of initial y boundary so this equation or what you want to study it may not be relevant for all x in R it may be that it is relevant only for x positive or maybe for x in some bounded interval a, b in such cases imagine the first case if it is meaningful only in R plus then x equal to 0 is a boundary of R plus right boundary point and if you are studying x in a, b then there are 2 boundary points x equal to a and x equal to b at these points so you need to prescribe some conditions there that is what is called initial boundary value problem. Now let us look at a picture illustration so this is the case where the domain is R plus x belongs to R plus therefore you have boundary at x equal to 0 so you prescribe this g of course initial condition you prescribe. Now when you are doing initial boundary value problem when x belongs to this interval a, b you have to give initial conditions and also give boundary conditions. So now let us look at 3 examples 3 guiding examples for us Cauchy problems because you can understand entire theory by using these examples. Before that let us ask the following question does every Cauchy problem have a solution good question this question sounds like a very familiar question that we asked earlier does every problem have a solution this is not the first time we came across such a question it was asked many times and many times we even had a complete answer for example system of linear equations where a, x equal to b varies a square matrix there we understand completely and what conditions you have existence what condition you have uniqueness and non-existence. Now let us ask the same question in some few well known situations for us before solutions of polynomial equations in one real variable what about that a polynomial may not have a solution polynomial equation x square plus 1 equal to 0 has no real solution it can have exactly one solution let us say x minus 1 equal to 0 or x into x square plus 1 equal to 0 they have exactly one solution then you can have finitely many solutions let us take very simple x minus 2 into x minus 3 equal to 0 right any finite number let us say a 1 a 2 a n you write x minus a 1 into x minus a 2 into x minus a n equal to 0 precisely these are the solutions a 1 a 2 a n but you cannot have infinitely many solutions for polynomial equations that is because of fundamental theorem of algebra it says given a non-constant polynomial it has exactly the same number of roots as that degree counting multiplicities in the complex plane. Therefore in real numbers may be less than or equal to that then solutions to initial value problems for ODE is this is where we have seen Picard's theorem and Piano's theorem in Piano's theorem if the right hand side let us say y dash equal to f x y let us write once this is the kind of equations for which we have a theory and this is the initial value problem if f was continuous we had Piano's theorem which said there is a solution. In addition if f is Lipschitz with respect to the y variable then Picard's theorem told us that it has a unique solution and when you do not have unique solution you can easily show that you have infinitely many solutions that is the situation about ODE so this is under sufficient conditions that we have proved these theorems. So if these sufficient conditions are not satisfied we have no idea what will happen anything may happen you may still have uniqueness even though the right hand side function f is not Lipschitz that is a possibility. Now solutions to boundary value problems we have a Bernstein theorem under some sufficient conditions it guarantees that a second order ODE boundary value problem posed for that will have a unique solution at least from the Dirichlet data okay. This if you do not if you are not studied BVPs you can look at any good book on ordinary differential equations you will find. Then solutions to transcendental equation this is the most difficult one you know it is not like polynomials so you have to worry with each separate equation is a different story there are once again sufficient conditions okay. Now let us see what are the examples so the examples we are going to see are of Cauchy problems in these 3 examples which demonstrate all 3 possibilities in this case what are the possibilities you have a unique solution infinitely many solutions and no solution. Now of course another option is the finitely many solutions which I have not thought about let us take that up in a tutorial session that discussion. So what does this mean it means that we do not expect existence uniqueness theorem for free for any first order PDE we do not expect so only under some special assumptions in other words sufficient conditions we can expect such theorems which was true in all our earlier cases let us say in the ODE initial value problems or boundary value problems that is a true. Okay let us look at a very simple first order PDE it is Ux equal to Cu, C is a constant we will look at a special constants later C equal to 1 and minus 1 later on in the case of initial boundary value problems here C can be any number real number so it is posed for X in R and T in R you do not see T in the problem so it is actually ODE and from our ODE knowledge we can solve this any solution must satisfy this equation Uxt equal to U0T into E power Cx. Now let us look at 3 Cauchy data what are they first one is U of 0T equal to T second is Ux0 and here also Ux0 that means initial Cauchy data is given here on T axis here in other 2 cases given on X axis. Now using this expression we will get this U0T you wanted T is supplied so T into E power Cx that is a solution in this case Ux0 that is T equal to 0 that is U00 into E power Cx but U00 must be 1 therefore any function of T such that T of 0 is 1 we will do of job it will be a solution Tt into C E power Cx is a solution whenever T0 equal to 1. Here if U is a solution then what is expected is sin X must be a multiple of E power Cx that is impossible because sin X and E power Cx are always linearly independent as functions on any interval that you consider. So message is that Cauchy data 1 unique solution Cauchy data 2 and 3 one case you have infinitely many solutions for Cauchy data 3 there are no solutions so 3 possibilities have been exhibited here. Now the why the Cauchy problems behave differently PDE was the same Ux equal to Cu then what is different U was prescribed on T axis we had uniqueness and when U is prescribed on X axis either we had infinitely many solutions or no solutions therefore we ask the following question is T axis something special for the equation or is it X axis which is special for this equation the reason for the first question is because we have uniqueness second question is we have many possibilities so who is really special here because special people have to be taken care with a lot of special care we will see that later on. So the answers we will see later as we study the first order PDEs. Now let us look at a simple first order PDE for which we pose IBVP and a few observations we will not be dealing with this further in our course. So the fact is that general solution of Ud plus Cu X equal to 0 is given by Uxt equal to Phi of X minus Ct where Phi is any C1 function which is arbitrary of course what is that T equal to 0 Ux 0 becomes Phi X. So this Phi is nothing but the initial solution initial condition. Now initial boundary value problem as I told you there will be a PDE posed in the first quadrant Xt X positive T positive initial condition is given and there is only one boundary point namely X equal to 0 therefore U of 0 T is also prescribed. Now Uxt equal to F of X minus Ct because see here we said Phi is initial data right. So therefore here also we have initial data so how much we can this determines the solution. So F of X minus Ct will be a solution surely no problem but whenever X minus Ct belongs to the domain of F, F is defined only for greater than or equal to 0 therefore this determines the solution whenever X is bigger than or equal to Ct we will see a picture soon. Now on the line X minus Ct the solution U will be constant it will be F of K X minus Ct equal to K K is greater than or equal to 0 therefore this is formula is applies and Uxt equal to Fk it is constant. Now let us look at the case where C equal to 1 this is the picture. So this is Ux 0 Fx this is U0 Tgt and U in this region in this region this the line X equal to T is in dots to the right side of that this region corresponds to X bigger than or equal to T there F of X minus Ct similarly you can show that here it will be G of T minus X that will be solution. So on this line T minus X equal to K solution will be Gk and on this line X minus Ct equal to K solution will be F of K. So initial condition determines U in this region and boundary condition determines in this region. Now what about X line X equal to T what will happen? So let us write the formula this is the formula that we said now if this has to be a solution it has to be differentiable the first question is is it differentiable? So this is a candidate solution we say because we have derived certain possible formula for the solution that is why we propose a candidate we need to verify that it is indeed a solution. So that requires that Fg must be C1 functions and compatibility of the data is needed F0 must be equal to G0 F dash 0 equal to minus G dash 0. How do you get these compatibility conditions? You simply ask at points where X equal to T write down the differentiability property of U of Xt that will tell you that this condition has to be met otherwise it would not be differentiable. So once you have these compatibility conditions that implies that U has U is a C1 function and IBVP has a unique solution. Now the Kc equal to minus 1 is totally different now the line is X plus T equal to constant this line touches both the boundary as well as the initial data. So and once again F must be constant on this line solution must be constant therefore this line touches at K0 so it should the value on this line should be Fk and it touches here at 0 K so the answer must be the solution must be G of K. So on this line the solution is both F of K and G of K what if F of K is not equal to G of K there is a problem. This means that G cannot be prescribed independent of F. This initial boundary value problem has no solution in general but there are some generalized notions of solutions where under some other condition they admit solutions. So we will not be discussing such generalized notions this is in the normal notion of a solution there is no solution that is very clear from the formulae. So let us summarize what we did today we have derived a first order PDE to model traffic flow and also modified to account for sources and sinks namely entries and exits in the express way and we highlighted the roles of balance law and constitutive relations in deriving mathematical models. This is how models in let us say continuum mechanics will be derived these are the two things which are at the heart of any modeling and then through three examples we understood that number of solutions to a Cauchy problem depends on the relation of the PDE to the datum curve and then we saw IBVP problems need not have solutions. So thank you.