 Welcome to this lecture on well-posedness of Cauchy problem for wave equation. So far we have solved the non-homogeneous Cauchy problem for non-homogeneous wave equation. Now we are going to show that this problem is well-posed which we are going to define what is the meaning of well-posedness and then prove. So the outline for this lecture is as follows. First we introduce the concept or the notion of well-posed problems in the sense of Hadamard. Then we discuss about existence of solutions, uniqueness of solutions and stability of solutions. These 3 properties are what are required for a problem to well-posed in the sense of Hadamard. So let us discuss what is called properly posed questions. Imagine you are taking an exam with multiple choice questions. Recall that for an MCQ type question you are given lot of options usually 4 let us say 4 options are given and exactly one of them is correct that is what is called multiple choice questions. Suppose that you found out 3 options are incorrect what will you do? Simply choose the remaining option as your answer without even looking at it. Why does it work? It works because you are told that one of the 4 options is correct. You have figured out that 3 of them are incorrect therefore what is remaining must be correct. So existence of an answer is guaranteed. Suppose you found out one option which is correct what will you do? Nothing else. Simply mark the correct option as your answer without even looking at rest of the options. Why does it work? It works because you are told that only one of the 4 options is correct. Uniqueness of an answer is guaranteed. So for any question mathematical or otherwise it is desirable to know the meaning of an answer that will tell you when do you know that you have question is actually answered. So when do you say that question has been answered and desirable to have the existence of an answer. Having defined what is the meaning of an answer it is desirable to have the existence of an answer. There could be questions without answers that means we understand what is the meaning of an answer to a question but that question does not have an answer. For example real roots for the quadratic equation x square plus 1 equal to 0. What do you mean by real root for this equation? It is a real number alpha such that alpha square plus 1 equal to 0 very clear. But there is no real number which satisfies alpha square plus 1 equal to 0. That means answer concept of answer is defined we understand what is the meaning of an answer but this question does not have an answer. And also desirable to have a unique answer. So in the case of multiple answers imagine you have more than one answer one feels that the question was not tightly posed or correctly posed. So please ignore this last comment for now. Imagine that I have not mentioned this part. So these are the three conditions. First thing is we should know meaning of an answer. It is desirable to have the existence of an answer desirable to have a unique answer. From here the Hadamard concept of properly posed questions will be stated. Now if you go back to questions regarding physical systems they are posed in terms of the corresponding mathematical models. In modeling lot of approximations are in model. Some are visible some are not visible. Even if the model is exact in order to answer related questions we need to rely on measurements. Somewhere we have to give input of the data and ask what is the output of the system. System is already modeled exactly using your equations mathematical models. But measurements are definitely approximate. Errors will be made in measurements. Therefore it is desirable that the inferences coming out of such approximate measurements or observations are closer to the reality. That is closer situation where such errors have not been made. Thus it is desirable that answers to questions do not change abruptly when the data in the question changes slightly. So well posed problems in the sense of Hadamard. A mathematical problem is said to be well posed or properly posed in the sense of Hadamard. So Hadamard has given this in French translation some people translated as well posed some people translated as properly posed. If the following requirements are met what are the requirements? Existence the problem should have at least one solution. Uniqueness the problem should have at most one solution. Continuous dependence the solution depends continuously on the data that are present in the problem. Note before asking if a mathematical problem is well posed one needs to define the meaning of solution to the problem. Only when it is defined we can ask whether you solution is exists or not unique or not etc. A remark on the well posed problems once the concept of solution is defined existence and uniqueness requirements are well understood well defined. However the third requirement namely continuous dependence is still not defined completely. Why? It is stated in terms of word continuity which is a topological property. Since the solutions and data belong to function spaces one needs to define ways of measuring distances between functions in terms of which continuity will be understood. One needs to identify such metrics which are relevant to the problem. A given mathematical problem may fulfill the requirement of continuous dependence with respect to one set of norms and may not satisfy this requirement with a different set of norms. Thus the requirement of continuous dependence is delicate. Continuous dependence requirement is also called stability requirement sometimes particularly in the problems where there is a time variable involved people call it stability requirement. So let us recall the Cauchy problem for wave equation. It is given functions phi, psi and f defined on the appropriate domains. Cauchy problem is to find a solution to the D'Alembertian operator equal to f and satisfying the two initial conditions. So rest of this lecture is devoted to proving the well posedness of this problem on the domain rd cross 0 t I will point out where exactly this is required for a fixed t positive. If you see here it is rd cross 0 infinity but if this result of well posedness will hold on rd cross 0 t only t cannot be made infinity. So proving well posedness of Cauchy problem we treat wave equations in 1, 2, 3 dimensions simultaneously that means we cover existence for all the 3 dimensions then we go to uniqueness and then we go to the stability results. So we use the word Cauchy problem without mentioning the D value. It is understood depending on the context. What is the meaning of solution because that is the first thing we have to do define before saying the problem is well posed in the sense of Adamard is to define the notion of solution to the problem that is the classical solution, notion of classical solution. We have derived the classical solution to the non-homogeneous Cauchy problem in the previous lectures. So well posedness of the Cauchy problem will be proved for this domain rd cross 0 t. The proof also suggests that we cannot expect such a result on the domain rd cross 0 infinity. So what are the steps involved in proving well posedness of Cauchy problem? First thing is existence. It is already shown that Cauchy problem admits classical solutions in the earlier lectures on the domain rd cross 0 infinity. We will recall those results uniqueness, we have not proved uniqueness we will prove that in this lecture. And then we go to the continuous dependence using the formula for solutions like D'Alembert formula or Poisson Kirchhoff formula in dimension 2 and 3, D'Alembert in dimension 1. We establish stability estimate in each of these dimensions d equal to 1, 2 and 3. So for this it is necessary to work with the domain rd cross 0 t. So it is for the stability estimate that we need to work with finite time. Precise hypothesis on the data phi psi f will be mentioned in the respective results. So let us go to existence of solutions. Here I am just going to recall what we have done in the earlier lectures. Solution to the full Cauchy problem sometimes I call this as non-homogeneous Cauchy problem in 1D. So phi should be c2 of r, psi should be c1 of r, f should be continuous and r cross 0 infinity and fx should be continuous on r cross 0 infinity. This is the formula that we have obtained reference is lecture 4.2 and lecture 4.7. Now in two dimensions the hypothesis required is here I will not read very fully but let us say phi and psi should be c3 and c2 respectively and f gradient f and secondary way to suffer should be continuous on r cross 0 infinity. And this is the formula that we have obtained reference lecture 4.6 and 4.7. So this represents a classical solution to the Cauchy problem. Now we may also write it on these domains d01 disc of radius 1 center at the origin because we are going to use this formula in proving the stability estimate that is why I have written down this particular formula. How about in 3D a similar hypothesis as in 2D and this is a formula and reference lecture 4.5 and lecture 4.7 that is where we have derived and shown that this is a classical solution to the non-homogeneous Cauchy problem in 3 dimensions. So now let us discuss uniqueness of solutions. So general idea for establishing uniqueness always is like this. To any problem if you want to show it has uniqueness you just take solutions u and v consider the difference it works usually in the linear situations even in the non-linear situations one considers that and then try to show that that difference is 0. Okay let u and v be solutions to the non-homogeneous Cauchy problem then the difference let us call w u minus v it is a solution to the following homogeneous Cauchy problem what is that homogeneous wave equation and 0 initial data because both u and v satisfy initial data phi and psi for initial displacement and velocity respectively difference will be 0 because of the linearity the problem is linear delambration operator is linear and value at a point derivative value at a point these are all linear operations. Now showing uniqueness means what I wanted to show that w is 0 so therefore showing uniqueness to the non-homogeneous Cauchy problem is same as showing that this homogeneous Cauchy problem everything is homogeneous with 0 initial data has 0 of course 0 is a solution very clear substitute w equal to 0 it satisfies but we have to show 0 is the only solution that sometimes the 0 solution is also called trivial solution 0 solution is the only solution to the homogeneous Cauchy problem which is here therefore we will concentrate on showing only this that the solution of such a homogeneous Cauchy problem with 0 initial data the only solution is the 0 solution thereby establishing uniqueness of solutions to the non-homogeneous Cauchy problem. Let us move to the one dimension recall from lecture 4.2 we solved the Cauchy problem for homogeneous equation first how did we do it was written in terms of characteristic coordinates and obtained its general solution as a result general solution was obtained in xt coordinates after that we use a Cauchy data and we derived the D'Alembert formula thereby we established that any classical solution to the Cauchy problem must be given by D'Alembert formula thus solution to the homogeneous Cauchy problem is the 0 function because for the Cauchy problem with homogeneous wave equation and 0 initial data 0 is already known to be solution advantage of uniqueness is that that is the only solution this completes the proof of uniqueness in one space dimension in other words we made no compromises in deriving the D'Alembert formula now let us move on to uniqueness in three space dimensions we remember this is the order in which we solve the Cauchy problems also first we solve dimension 1 then we solve dimension 3 then we used Hadamard's method of descent to solve in two dimensions okay so recall from lectures 4.4 and 4.5 mw of rho t it is the spherical means associated to the function w if w is a solution to the homogeneous Cauchy problem in D equal to 3 then this l of rho t given defined by rho times mw of rho t this is what we did in lecture 4.5 this l of rho t satisfies a one-dimensional wave equation okay and if w satisfies the homogeneous Cauchy problem then l also satisfies the homogeneous Cauchy problem in one dimension and there we have just shown on the previous slide the trivial solution is only solution therefore l of rho t must be the 0 function once l of rho t is 0 mw of rho t must be 0 once mw of rho t is 0 it means all spherical averages are 0 by l y sm laman spherical means w itself must be 0 which means we have got uniqueness therefore solutions to non-homogeneous Cauchy problems in 3D are unique now let us move on to two space dimensions recall from lecture 4.6 if w is a solution to the homogeneous Cauchy problem for D equal to 2 then w is also a solution to the homogeneous Cauchy problem for D equal to 3 we have just proved uniqueness therefore w must be 0 this proves uniqueness result for non-homogeneous Cauchy problem in two space dimensions. Now let us move on to stability of solutions what we mean by that we are going to state in the form of result let us do one space dimension first stability result hypothesis is consider this Cauchy problem of course we have to write f i psi the hypothesis on them so that we have a classical solution to this problem that is what is going to be hypothesis of course we have introduced a t so t should be positive you will see this t will play a crucial role the last step of the proof you will see exactly where and how the t plays a role so assume the hypothesis on f phi psi which are required to have a classical solution to this problem so conclusion is given epsilon positive now we are going to write about the continuity right so given epsilon positive there is a delta such that whenever two data that is f 1 phi 1 psi 1 this one data set this is another data f 2 phi 2 c 2 whenever these two are at a distance of utmost delta okay I am using very loosely the word distance mod phi 1 x minus phi 2 x is always less than delta for every x similarly mod psi 1 x minus psi 2 x is less than delta for every x and mod f 1 x t minus f 2 x t is less than delta for every x t in r cross 0 t that happens then the corresponding solutions u 1 denote solution with this initial data u 2 denotes with this as the not initial delta f 2 is the source term phi 2 psi 2 as the initial displacement and velocity respectively so u 1 and u 2 they satisfy what is called the stability estimate mod u 1 x t minus u 2 x t is less than epsilon for every x t in r cross 0 t so this tells that if the data two data sets f 1 phi 1 psi 1 and f 2 phi 2 psi 2 are sufficiently close then solutions will remain arbitrarily close the close that you want you prescribe this epsilon then such a delta exists this is a continuity kind of requirement okay solution to the non-homogeneous Cauchy problem is given by this formula we know this and what is the stability estimate it is in terms of u 1 minus u 2 so we write this formula for u 1 u 2 and subtract and get u 1 minus u 2 that is idea so let u 1 u 2 be solutions to Cauchy problem with not really Cauchy data this is the Cauchy data this is the source term similarly phi 2 psi 2 is Cauchy data and f 2 is a source term subtracting the formula for u 1 u 2 we get this okay just substituted and then subtracted now notice how the things look phi 1 minus phi 2 phi 1 minus phi 2 psi 1 minus psi 2 f 1 minus f 2 what we have to show is that there is a delta such that whenever these differences are at most delta in modulus this can be made less than epsilon that is what we want to show so let us apply the triangle inequality and we get this modulus of the LHS less than or equal to modulus modulus of the sum is less than equal to sum of the modulus so apply this and modulus of the integral is less than or equal to integral of the modulus using all that we get this from the previous slide okay now if there is a delta like that we are going to find the delta but if at all there is a delta with this thing then I will go and see what consequence it has to this estimate so I get mod u 1 minus u 2 less than or equal to this is less than delta so delta by 2 this is less than delta no matter what is argument so delta by 2 plus delta by 2 plus 1 by 2 c x minus c t this is also less than delta delta comes outside then you get length of this interval which is 2 c t similarly here this is less than delta it comes outside then as you know this is the triangle so the triangle area will come this is the first phi 1 minus phi 2 at x minus c t this is a second term phi 1 minus phi 2 at x plus c t this integral term with psi this other integral term with f please do this computation very simple simplifying we get this okay now what is our idea we want to make this less than epsilon therefore given epsilon I want to make this less than epsilon can I choose a delta so that this is less than epsilon yes choose delta to be less than epsilon by 1 plus t plus t square then above computation shows but mod u 1 x t minus u 2 x t is less than epsilon okay now comes a comment if t is higher and higher this delta you can smaller and smaller imagine t is infinity so loosely speaking this is 0 that means you would not be able to choose this delta that means it gives you an idea that if you consider the Cauchy problem r cross 0 infinity we may not have the stability estimate okay that means there may be there will be or there may be the data sets f 1 phi 1 psi 1 which are close to f 2 phi 2 psi 2 but solutions are not so that I leave it for you as an exercise to think about it how to get that given that we are dealing with a with a linear equation you may consider one of the data set to be 0 0 0 okay so this completes the proof of the theorem so we measure distances between data and distance between solutions in the kind of uniform sense okay I have put this in quotes because it is exactly not uniform sense but some kind of uniform sense it is not uniform metric or if you know it is not supremum now no because on the spaces we are considering they are like c of r cross 0 0 t okay with respect to r you may not have supremum now for any continuous function imagine f x equal to x of course it is in c of r cross 0 infinity supremum known does not make sense okay that is why I am not using that word here that is why I have put quotes here uniform in the sense for every x distance between phi 1 and phi 2 is less than delta that is what I mean so one can also consider questions of stability with respect to other notions of distances between functions in such case the same question problem may be either stable or unstable we have no idea anything can happen because if you have studied functional analysis you know that an infinite dimensional norm in your spaces two norms may not be equivalent that is a problem so topologies could be different so thus there is a need to state the result in clear mathematical terms this is always the case if you think you have proved some result you should be able to state it very cleanly many very clearly it can run into any number of pages but it should be very clearly stated that is when we know what we have actually proved instead of expressing it in some plain sentences let us move on to the two space dimension stability of solutions result is going to look exactly as before instead of 1d wave equation you have 2d wave equation f e and psi so assume f belongs to continuous grad f is continuous and dx 2 is continuous assume phi is c 3 psi is c 2 so that we have a classical solution to this problem so conclusion given epsilon there is a delta such that whenever the data triples are at most of a distance in some sense of delta uniformly with respect to x or uniformly with respect to x t in this case then the corresponding solutions will satisfy this stability estimate the idea proof is the exactly the same write down the expression for u 1 expression for u 2 subtract apply triangle inequality and try to show this so this is a formula for you which we know this is this is a convenient formula that is why we are using this formula okay let u 1 u 2 be solutions for the data f 1 phi 1 psi 1 f 2 phi 2 psi 2 subtract the formula we get this expression now apply take the modulus on both sides and apply triangle inequality here there are 1 2 3 4 terms on the RHS just for convenience let us give some names a 1 a 2 a 3 a 4 so therefore modulus of u 1 minus u 2 of x t is less than or equal to mod a 1 plus mod a 2 plus mod a 3 plus mod a 4 and what is mod a 1 is less than or equal to the modulus inside and phi 1 minus phi 2 is always less than delta that is our assumption that comes out delta by 2 pi and what is left is integral d 0 1 dz by root 1 minus norm z square similarly here you can bring things outside okay so what we get is mod a 1 plus mod a 2 first 3 terms I am considering first term will be delta by 2 pi as we already saw into this integral second term will be less than or equal to c t delta by 2 pi into this integral term third term mod a 3 is less than or equal to delta by 2 pi into t into this integral that is what you will get so we have to compute what this integral is use polar coordinates and then this evaluates to 2 pi so therefore first 3 terms are less than or equal to delta into 1 plus c t plus t that 2 pi cancels with this 2 pi now one more term is there proceeding in a similar manner the fourth term will satisfy this estimate mod a 4 is less than or equal to 2 pi c delta t square combining all the estimates we get this expression now can I make this less than epsilon yes choose delta to be less than epsilon divided by this quantity so we will get this so this completes the proof of the theorem now move on to 3 space dimensions here we just state the result exactly same as before here t belongs to 0 t t belongs to 0 t the assumptions on f phi psi so that this problem has a classical solution so given epsilon positive there is a delta positive so is there whenever you have data troubles with the required regularity or smoothness as on the previous slide and satisfying this that they are uniformly at a distance of delta at most distance delta and also f 1 f 2 distance is less than delta uniformly for x t in r 3 cross 0 t corresponding solutions even you to satisfy this estimate the proof is exactly the same as in the d equal to k s and we skip it so let us summarize what we did for the non-homogeneous Cauchy problems uniqueness of solutions was proved we are also going to see another proof of uniqueness later on continuous dependence of solutions on the data triple was proved in the form of stability estimates this concludes the discussion on Cauchy problem for wave equation in full space that is x belongs to r r 2 and r 3 in the next few lectures we discuss initial boundary value problems they arise when the wave equation is posed not in the full space r d but on subsets of r d in fact we are going to study a subsets of r only mostly in one space dimension thank you