 Welcome. In this lecture, we are going to discuss Cauchy problem for the wave equation in d space dimensions. In fact, we are not going to discuss fully. What we are going to do is we are going to reduce this Cauchy problem in d space dimensions to a Cauchy problem for a PDE which is having just a 2 independent variables that we do a biospherical means. Outline for today's lecture is we start with statement of the Cauchy problem for the wave equation in d space dimensions. Then we introduce spherical means and a formula known as Darbou formula associated to that and then we reduce the Cauchy problem for the wave equation to an equivalent Cauchy problem. So, Cauchy problem for d dimensional wave equation reads as follows. Given functions phi and psi solve the homogeneous wave equation in d space dimensions as you see x 1, x 2, x d are the space dimensions are the space variables. And u of x 0 is phi x, ut of x 0 equal to psi x. We are not at stating what is the regularity or the smoothness of the functions phi and psi required. We will see them later on. So, key idea to solve this Cauchy problem is reduce the Cauchy problem to an equivalent Cauchy problem in one space dimension advantage the number of space dimensions drop from d to 1 and the tool is method of spherical means. So, when d equal to 3 the new Cauchy problem which we have obtained it can be transformed to a one-dimensional wave equation Cauchy problem for a one-dimensional wave equation of course after a change of dependent variable which is straight forward and then we are on with the Dalambert formula which gives a solution to the Cauchy problem for the one-dimensional wave equation and we can go back and retrieve the solution to the three-dimensional wave equation the Cauchy problems solution from the Dalambert formula that we get here. So, in this lecture we confine ourselves to deriving the equivalent Cauchy problem valid for all space dimensions. So, let us see a few computations and notations involving multiple integrals. So, let us start with notations. So, we deal with functions of d real variables not necessarily d equal to 3 Euclidean norm of a vector x is denoted by norm x that is a notation and that is given by a square root of x 1 square plus x 2 square plus up to plus x d square the usual length of the vector x. B of x comma rho it denotes the open ball of radius rho of course radius has to be positive and with center at x. So, B of x comma rho is set of all y in Rd whose distance from x is at most rho that is norm x minus y is less than rho. S of x comma rho denotes this field having radius rho and center at x that is S of x rho is y in Rd such that norm x minus y equal to rho. The usage of the word sphere is slightly confusing because from our high school by sphere what we mean is a sphere which is in this case actually the ball union the sphere the boundary of the ball is the sphere both of them if you take the union you get the so called solid sphere. So, what we mean by sphere here is actually the hollow sphere hollow sphere is called sphere in our course. In fact, that is so in any advanced mathematical courses and solid sphere is the union of the ball and the sphere. So, you should not get confused. Now, omega d denotes the measure of the unit sphere S of 01 in Rd omega 2 for example, d equal to 2 omega 2 is a perimeter of the circle of radius 1 therefore it is 2 pi omega 3 is a surface area of the sphere of radius 1 and that is 4 pi. We also have this relation which connects the surface area of the unit sphere to surface area of sphere of any radius. So, this is a surface area of the sphere of radius rho that is rho power d minus 1 times the surface area of the unit sphere. d omega is used to denote the surface measure on S01 we are going to deal with various integrals on S01. So, we write d omega denotes the surface measure. Now, we use a generic notation d sigma to denote surface measure on any hyper surface in our context since we are dealing only with spheres we said d sigma is used to denote the surface measure on any sphere of course, it varies from sphere to sphere. In fact, it varies with the radius of the sphere. Now, let us look at the change of variables from so it is a mapping from B of x rho to B 01 y in B of x rho is mapped to y minus x by rho. So, the ball of radius rho center x will be going to the ball of radius 1 and center 0 under this mapping it is of course, it is an invertible mapping. So, given a function G defined on B of x rho we can define a function which is defined on B of 01 that is W W of z for z in B 01 by G of x plus rho z. So, a function defined here can be transferred to a function here and vice versa. So, G of y is given by W of y minus x by rho in terms of W. So, this equation gives rise to the following equations of course, chain rule. So, we are going to differentiate both sides with respect to y. So, gradient with respect to y is gradient of W with respect to z and gradient of the inside quantity which is y minus x by rho with respect to y that will give you a 1 by rho that is why rho inverse. And Laplacian consists of second order derivatives therefore, you get one more rho inverse. So, you have rho power minus 2 into Laplacian in z of W evaluated at y minus x by rho. So, we have to be as I told you we have to be very comfortable with using the chain rule particularly when there are change of variables involved. So, the measure dy is given by rho power d into dz in view of that and the equation that we have just derived Laplacian y G in terms of Laplacian z W what we get is this particular integral. Now, we are going to do change of variable z equal to y minus x by rho the domain of integration transforms to this b of 0 1 and the integrand will be rho power minus 2 into Laplacian z and the dy the measure becomes rho power d dz. So, pulling the terms outside what I have is rho power d minus 2 into integral over ball of radius 1 center 0 Laplacian z W. So, now from here we get this by integration by parts or Green's theorem. So, we use change of variables and integration by parts in these computations. Now, we also had one relation for the gradient under change of variables. So, using that the last equation that we obtained on the previous slide can be simplified expressed in terms of G. So, this integrand is precisely dou by dou rho of this quantity and dou by dou rho will come out of the integral now. So, this is the expression that we have got. So, this is what we have proved integral of the Laplacian G over b of x rho is given in terms of rho power d minus 1 dou by dou rho derivative of this quantity the integral on the unit sphere radius unit sphere with center origin. Now, we are in a position to introduce spherical means and then derive Darbou's formula. So, take a continuous function G defined on R d to R let that be a continuous function. The spherical mean of G denoted by mg, m for mean we are talking about mean of the function G because later on we are going to deal with means of at least three functions. One is mean of the solution u and mean of the initial data phi and psi. That is why we introduce G here not to get any confusion later on. So, mg of x comma rho it is a function defined on R d cross 0 infinity taking values in R. So, mg of x rho is defined by integrate G over the sphere S x rho and divide with the measure of S x rho that is why it is called mean. And d sigma is a surface measure on this sphere and what is the measure of S x rho that is precisely rho power d minus 1 into omega d this we have already stated. Spherical means means spherical averages. So, this formula which defines spherical means is integrating the function G over the sphere and compares it with the surface area of the sphere. Recall how arithmetic means are defined. This is exactly similar. So, spherical means are averages of the function over spheres. Now, we have a question why are we considering spherical means? Are there any advantages to it? Knowing all the spherical means of a function is same as or is equivalent to knowing the function itself and importantly function can be retrieved from its spherical means that is idea. We will state more precisely results which say which assert function can be retrieved from spherical means. We will do that. So, now another question is are we complicating by introducing an additional variable rho? Not at all. Once we fix x the spherical means are functions of only one variable namely rho. So, this will help us in reducing the wave equation in d dimensions to an equation with a time variable and one more independent variable. We will see that. Let us look at this computation where we are integrating G over the sphere sx rho. That by Green's theorem is equal to this. So, please compute this quantity on the right hand side. You will get the quantity on the left hand side. Now, why are we doing this? We will see at the end of this computation the reason for doing this. Now, this we are now changing the variable where ball of x rho is now becoming ball of 0 1 and the measure dy naturally becomes rho power d dz and the derivative dou by dou yi is rho inverse dou by dou zi. We have already established these facts. So, finally we have an integral on ball of 0 1. Now, we will write this using once again integration by parts of Green's theorem and express this as an integral on the sphere s 0 1. So, we end up with this expression and if you notice this is does not depend on i. So, it can come out what I have a summation i equal to 1 to d zi square. Where is z? Z is on the sphere. Therefore, norm z is 1 and summation i equal to 1 to d zi square is 1. Therefore, we get this expression. So, we have got integral on s x rho. We have converted that to an integral on s 0 1. So, we did change of variables in spheres. So, in view of the computation on the previous slide, what was it about? We have just answered it is something to do with change of variables on spheres. Usually, we learn change of variable theorem for open sets in Rd. So, that means we know how to handle open sets in Rd whereas s x rho is not an open set in Rd. So, directly if you write this equal to the last line, there are questions and this is what is establishing that this is indeed true. So, we can behave like yes, I know how to do change variables even for spheres. So, that is about it. Now, this is the definition of mg of x, rho. Now, that we have an alternate expression which is 1 by omega d integral of g of x plus rho nu d omega integral over norm nu equal to 1. So, in view of the definition of spherical means which is actually mg equal to this. Now, we have established it is equal to this. So, we have a new formula for mg. So, spherical means was defined by the formula mg equal to 1 by measure of s x rho into integral over s x rho of gy d sigma and we obtained another expression. Why are we doing this? What is the advantage of the second formula over the first one? The answer is observe the domains of integration in this formula. What is the domain here s x rho? What is the domain here norm nu equal to 1? What is this function mg of x comma rho? So, if I want to differentiate either with respect to s x or with respect to rho here, it is difficult because the domain itself is varying with x and rho whereas in this expression it does not depend on x it does not depend on rho that is the advantage we have. So, we have a lemma on spherical means we will often be using we calling it l o s m lemma on spherical means hypothesis g is a continuous function and let mg be defined by formula the spherical means of g contribution 1 sometimes we may call l o s m 1 what is it? mg can be extended to the domain r d cross r such that for each fixed x as a function of rho it is an even function. Currently it is defined only for rho positive because we needed the ball to have a positive radius. So, let k belongs to n if g is a ck function on r d then so is this function x rho mapping to mg of x rho it is a function defined on r d cross r. Note by one conclusion 1 we have already extended the function mg of x rho for rho belonging to r and the function g can be recovered from the spherical means this is the most important property what we mean by that you take mg of x rho and then take limit as rho goes to 0 you get g of x and this happens for every x. Now let us prove conclusion 1 observe this equality holds why because the unit sphere is invariant under the mapping nu going to minus nu. So, therefore these integrals are same therefore mg which we know is equal to this is now also equal to this quantity. So, in other words mg as a function of rho whether you put plus rho or minus rho it is the same that means the function is even. So, we can use the same formula to extend mg of x rho even for rho negative. So, we can do this for rho negative rho positive we started defining spherical means for rho positive we have now extended for rho negative what about 0. So, 0 we just take it to be g x mg of x 0 equal g x this is also consistent with the third assertion of this lemma then it is defined for all rho. So, just summarize we have defined mg of x rho for rho positive then we observe that if you replace rho with minus rho the integral does not change therefore we can extend mg of x rho even for negative rho and for rho equal to 0 we define by mg of x 0 equal to g x. So, let us go to the proof of conclusion 2 if g is a ck function in rd defined in rd then from this formula mg of x rho equal to 1 by omega d integral over norm nu equal to 1 g of x plus rho nu d omega notice that x and rho appear inside the argument of this function g therefore if g is ck then mg of x rho will be ck also. So, it follows that this function x rho going to mg x rho belongs to ck of rd cross r how we have to justify the interchange of the derivative and the integration and that is an exercise in analysis now proof of conclusion 3 let us rephrase the conclusion this is limit rho goes to 0 this is mg of x rho that is equal to g x this is what we want to show. So, how do you show this start estimating the difference between this quantity and g x. So, let epsilon be given positively given let us estimate this quantity whose limit you want to find you want to show that limit is g x therefore estimate the distance now g x can be written as g x can be written as 1 by omega d integral norm nu equal to 1 because omega d is after all the measure of the set norm nu equal to 1 therefore we can do that of course g itself does not depend on nu so we can put it inside. Now we take the modulus inside the integral so modulus of the integral is less than integral of the modulus and we get this now this suggests something to be used the property of g here g is continuous at x therefore this can be made small as a result this can be made small as a result this can be made small and we achieve what we want to show that is the idea since g is continuous at x there is a delta such that whenever y is delta close to x g y minus g x is less than epsilon in modulus therefore we are not interested in arbitrary y but we are interested in y which is looking like x plus rho nu x plus rho nu where nu is having unit norm. So therefore for mod rho less than delta this becomes mod g of x plus rho nu minus g x becomes less than epsilon. So now conclusion 3 follows from this estimate because this is less than epsilon so the RHS here is less than or equal to epsilon by omega d integral norm nu equal to 1 d omega which will give you omega d so omega d omega d cancels and you get epsilon. So the conclusion 3 follows now let us prove Darbu's lemma for g in C2 of Rd the Darbu formula holds what is the Darbu formula it exists Laplacian in the x variable of mg of x rho is this quantity. So Laplace x sub x stands for the Laplacian operator the variables x 1 up to x d which is here. Actually if you are looking at the Laplace equation we have not had looked at but Laplace equation and you want to look at solutions which depend only on the radius R then that you end up with this operator. So this is actually the radial Laplacian other angular parts are ignored I am not looking at solutions which depend on the angles therefore only the radial part this is that. Recall that we have proved this Laplacian on B of x rho is given by this quantity this we already proved in the beginning. So this quantity is actually this with a omega d 1 by omega d missing so you put omega d so you get this multiply both sides with 1 by omega d you have this equality. Now express this integral on the ball using polar coordinates R and d omega so you have this expression. Now differentiate both sides with respect to rho so LHS is simply this and RHS by fundamental theorem of calculus is the integrand evaluated at rho which is this. Now we can expand these derivatives which are here and we get this now if you cancel rho power d minus 1 on both LHS and RHS what we get is precisely that the Arbou formula. Now let us reduce the Cauchy problem for the wave equation to an equivalent Cauchy problem. So for that we have to introduce spherical means associated to the solution or the unknown function u. So let u be a c2 function be a solution to the Cauchy problem for d dimensional wave equation for each fixed t and x define the spherical means m u x t rho earlier when we were dealing with spherical means there was no variable called t therefore it was missing we had only x and rho but now we have to include that because we are trying to solve a wave equation now so m u of x t rho is exactly as before. Now apply Laplacian on both sides of the above equation what do you get? What we get is this is just applying Laplacian and writing so we need to really do here the Laplacian. Laplacian has gone inside here that is the change now u is a solution right. So Laplacian x u is utt by c2 because you use a solution to the homogeneous wave equation. So then we have this now bring dou 2 by dou t square outside and we have this. So what we have is a Laplacian m u is 1 by c2 dou 2 by dou t square m u. Now we have the statement about the equivalent Cauchy problems. So the following statements concerning a function u which is c2 of r d cross r are equivalent. What are those? u is a solution to the Cauchy problem for d dimensional wave equation and that is same as saying dou 2 by dou t square of m u is c square into this operator acting on m u and of course the initial data for m u and dou by dou t of m u. We will prove this very simple proof. We have done all the computations I will just indicate how to go about this proof. So statement 2 follows from statement 1 by direct computation using the definition of m u. We have done all the computations in view of the computations immediately preceding the statement of this lemma and the Darbou formula. Statement 1 follows from statement 2 in view of Darbou formula and conclusion 3 of LOSM that is the function can be retrieved from its spherical means. So you are urged to fill in the details of the proof. Very simple details you have just write down a few lines. All the computations are already there and I have given you how to go about here. So let us summarize what we did in this lecture. We have introduced spherical means for a continuous function. Found that point wise values of the function may be obtained from the knowledge of all spherical means. Cauchy problem for wave equation in d space dimensions has an equivalent formulation involving only 2 independent variables. Thus reducing the complexity of the problem. In the next few lectures Cauchy problems for d equal to 3 will be solved using the equivalent formulation. Further solution to the Cauchy problem for d equal to 2 will be obtained using the solution that we get for d equal to 3 case. So that is what is called Hadamard's method of descent. So thank you.