 Welcome, in this lecture we are going to discuss about the propagation studies for the wave equation and with this lecture we are going to end our discussion on wave equation. Outline for this lecture is as follows, first we discuss about propagation of confined disturbances we will mention what they are later and then propagation of singularities and we give a few results on decay of solutions without proof propagation of confined disturbances. So we are going to study propagation properties associated with the Cauchy problem where the wave equation is homogeneous and we have the Cauchy data phi and psi which is confined it means the functions phi and psi are compactly supported. As Cauchy data we choose functions phi and psi to be piecewise constant functions with such functions the computations are easy and transparent. We have to bear in mind that the corresponding solutions given by the formulae like D'Alembert Poisson Kirchhoff are to be interpreted as weak solutions or weaker solutions that is why I have put this them in quotes. Of course we have introduced the notion of weak solutions in the lecture 5.6. Considering piecewise constant Cauchy data will also help us in analyzing propagation of singularities in the Cauchy data. Let us see an example dimension 1, one dimensional wave equation psi is identically equal to 0 and phi is such that it is equal to 1 on the interval 0 1 0 otherwise outside the interval 0 1 function is 0. So it is a discontinuous function but piecewise constant function 0 up to 0 1 from 0 to 1 once again 0 from 1 to infinity that is the function phi. The support of phi is precisely 0 1. So we are assuming that psi is identically equal to 0. Therefore as we see the support of the Cauchy data is the interval 0 1 which is a compact set. Yeah, support of phi is 0 1 psi is identically equal to 0. Now solution to the Cauchy problem given by D'Alembert formulae is this fix a t naught we would like to find the support of u of x, t naught. So we want to study the support of the solution to the wave equation at a time t equal to t naught fix a t naught then what we need is u of x comma t naught. So it depends on x minus c t naught and x plus c t naught therefore we will find out what is phi of x minus c t naught and phi of x plus c t naught we can easily see that this formulae hold because phi is equal to 1 whenever the argument is in interval 0 1. The argument is in interval 0 1 means x is in this interval. Similarly here this argument is in interval 0 1 means x is in this interval. So note that u of x t naught is non-zero for all those x for which x is in this interval or this interval. Thus for sure u of x comma t naught is 0 when x is not in the interval minus c t naught comma 1 plus c t naught. In other words support of the function x mapping to u of x comma t naught is contained in the compact interval minus c t naught comma 1 plus c t naught. By the same reasoning the support of u of x t that is x going to u of x t is contained in the compact interval minus c t comma 1 plus c t. This example illustrates that if Cauchy data has compact support then the solution at every time instant will also have compact support. Now let us like another example where phi is 0 but psi will be there. In the example 1 psi was 0 in example 2 phi will be 0 and psi has compact support say 0 1. Now fix t in t naught then dilumbrate formula gives you this as a formula for the solution. Thus for sure u of x t naught equal to 0 when x is not in this interval this is minus c t naught this is 1 plus c t naught. So when x is not in this interval let us say x is here what happens x is bigger than 1 plus c t naught that means x minus c t naught is bigger than 1. What does this mean? If I take the interval 0 1 x minus c t naught is here. So where will be x plus c t naught this side. What does this mean? Psi is supported in the interval 0 1 and x minus c t naught x plus c t naught which is a domain of integration is not intersecting 0 1 therefore the integral will be 0. Similarly if x is here because if x is not in this interval means x is either to the right side of 1 plus c t naught like here or to the left side of 1 minus c minus c t naught. So if x is like this what does it mean? x plus c t naught is less than 0 what does it mean? x plus c t naught is here where will be x minus c t naught it will be here. Once again this interval on which we are integrating psi does not intersect with 0 1 therefore this we have this u of x t naught is 0. In other words the support of this function x going to u of x t naught is contained in this compact interval that is what we have drawn the diagram and shown. No decay of solution this is another property that there is no decay when we are dealing with one dimensional wave position. x equal to x naught what is the meaning of decay? You stand at a point x 0 and look at u of x naught comma t as t varies as t goes to infinity. So for t naught bigger than this quantity we have x naught minus c t naught to be less than 0 and x naught plus c t naught bigger than 1 therefore what will happen is that 0 and 1 are here x naught minus c t naught and x naught plus c t naught will be here. If you recall the Dallambert formula that will be an integral on this interval x naught minus c t naught x naught plus c t naught. So if t is bigger than this t naught x naught minus c t naught x naught plus c t naught this interval will always contain the interval 0 1 on which psi is supported therefore the solution is actually 0 to 1 psi s ds that integral part so 1 by 2 c will be there. So we have this this will be the solution let us see that. So for t bigger than or equal to t naught this is the formula that comes from the Dallambert formula. Now RHS is a constant because as I told you this integral is from x naught minus c t to x naught plus c t but x naught minus c t and x naught plus c t always contains 0 1 the moment t is bigger than or equal to t naught where t naught is equal to this or bigger than this whatever equal to we can set. So and it is nonzero if this is not 0 it all depends on what is the integral of psi on the interval 0 1 if that is not equal to 0 this is a nonzero constant for all t it means the solution does not becomes 0 or does not decay at it stays constant. So we are in a big for a big trouble if sound waves propagate according to 1 d wave equation because of this reason luckily they do not. So d equal to 3 straight away we do the example for dimension 3 I assume speed is 1 c equal to 1 and phi equal to 0. So solution to the Cauchy problem this we have worked out in an earlier lecture to be this. Now consider an x with norm x equal to 2 and analyze what is it it says here I fix an x with norm x equal to 2 and I want to study u of xt as t goes to infinity what is the behavior. From the formula on the last slide we get u of xt equal to 0 for t less than 1 and for t bigger than 3 the function t going to u of xt is increasing interval 1 2 decreasing in 2 3 and then becomes 0 after t greater than 3. This is what we have already learned in earlier lectures that in 3 dimension solutions for the wave equation there is a time up to which information has not reached that is up to 1 after that information has reached and after this information goes away 0. So this is the leading edge and trailing edge that is what we have seen even the pictures when we discussed Huygens principle we have discussed them. So this behavior is very different from that for d equal to 1 we have just seen as illustrated by example 2. So where the solution becomes non-zero after some time and remains constant that could be non-zero if integral of 0 to 1 of psi is not 0. Now d equal to 3 c equal to 1 psi equal to 0 this Cauchy problem we have solved solved in quotes maybe a weaker notion of solution we got a u to be like this. Once again consider an x with norm x equal to 2 from the expression for solution we get u is 0 of a t less than 1 it is function is decreasing in 1 3 in fact this please verify all this assertions by yourself by looking at the formula and then become 0 for t bigger than 3. Now let us discuss propagation of singularities singularities travel along characteristics for d equal to 1. Let u be a solution to the homogeneous wave equation assume that for a fixed time t 0 that means that at a fixed time t 0 the solution u is not a c 2 sometimes we just call smooth function at the point x 0 t 0 that means that formula for u has a trouble at x 0 t 0 u is given by this expression f of x minus c t plus g of x plus c t. u has a trouble at x 0 t 0 means f should have a trouble at x 0 minus c t 0 or else g should have a trouble at x 0 plus c t 0 or both. So this means either f is not smooth at x 0 minus c t 0 or g is not smooth at x 0 plus c t 0 why please justify this if they are smooth then you can conclude that u is smooth smooth if you think it is continuous it is continuous if it is differentiable it will be differentiable if you think it is c 2 it will be c 2. Now observe that there are two characteristic lines which pass through the point x 0 t 0 they recall that there are two families of characteristic lines for the wave equation one member each passes through this point x 0 t 0 namely x minus c t equal to x 0 minus c t 0 x plus c t equal to x 0 plus c t 0 these are two characteristic lines which pass through the point x 0 t 0. Now in view of this formula for u if f is not smooth at x 0 minus c t 0 then f will not be smooth at all those x and t such that x minus c t is equal to x 0 minus c t 0 after all it depends on whether f as a function of one variable what happens to at a particular location x 0 minus c t 0. So therefore whenever x minus c t is equal to x 0 minus c t 0 you have the same problem that is why it is not smooth. So if g is not smooth at x 0 plus c t 0 you will not be smooth at all points on this line x plus c t equal to x 0 plus c t 0 for the same reason this shows that the singularities in solutions to wave equation are traveling only along characteristics. Note that the nature of singularity also does not change that is what I said if f is not continuous you will not be continuous if f is not differentiable you will not be differentiable if f is not c 1 you will not be c 1 and so on. So the nature of singularity does not change. Let us look at what happens in higher dimensions let you be a radial solution to the Cauchy problem in 3D where the functions phi and psi are also radial what do you mean by radial? It means that the function depend only on the distance to the origin. So phi of x will be a function of norm x that is there exists f and g is such that phi x equal to f of norm x psi x equal to g of norm x. In other words phi and psi are constant at all points of any sphere with centered origin. Since phi and psi are functions of norm x only in other words it depends on the distance of x to the origin these functions are called radial functions because if you consider a sphere with radius norm x norm x is the radius that is why these are called radial functions. Let phi belongs to c 3 of r 3 and psi belongs to c 2 of r 3 these are the assumptions that we need so that this problem will have a classical solution given by Poisson Kirchhoff formula and that in turn means that f and g they are c 3 and c 2 on this interval 0 comma infinity. So we look for solutions u of xt which are radial that is u of xt is looking like u tilde of norm x comma t where u tilde is a function of r comma t r is in the interval 0 comma infinity and t is in 0 comma infinity once again. So note that u tilde solves the Cauchy problem which is given here it is very easy to derive in fact we have already done this before when we are trying to get solutions to the Cauchy problem. So this is in fact is what is called as radial Laplacian. Here f and g denote the extensions to r of the given f and g because the given f and g are defined only on the closed interval 0 comma open infinity whereas here we need r belongs to r because we are trying to pose a problem for r belongs to r and hence we extend the given f g to f g so we still use the same notations f g such that the extended functions are even functions and of course f is in c 3 and g is in c 2. This requires that f dash of 0 and f double dash of 0 is 0 similarly g dash of a 0 equal to 0. If f and g satisfy these conditions then we can do this extension as mentioned here in this point assume these conditions. Now defining v of r t equal to r into u tilde of r t we see that v satisfies the Cauchy problem for the 1D wave equation because under this change of the dependent variable the radial Laplacian will simply become dou 2 v by dou r square. So this is the exactly one dimensional wave equation for v and these are the Cauchy data v of r 0 is r f r dou v by dou t at r 0 is r times g of r. Now using D'Alembert formula we conclude that u tilde of r t is given by this formula. So of course D'Alembert formula gives you v of r t but once you know v of r t you know what is u tilde of r t you have to divide with r therefore I divide with r I get this formula for u tilde of r t. From now onwards assume that psi is identically equal to 0 this means that there is no g, g is identically equal to 0. So the formula simplifies in this special case to this formula u tilde of r t equal to 1 by 2 r into r minus t into f of r minus t plus r plus t into f of r plus t. Using L hospitals rule and that the function f is even we get u tilde of 0 comma t equal to limit of this quantity as r goes to 0 which using L hospital rule turns out to be f of t plus t times f prime of t. This is another illustration of loss of derivatives because we have assumed f is c 3 but now you have f dash so that means it is just c 2. So u tilde will only be c 2, u tilde of 0 t if it exists it will be only c 2 function. So we have lost the derivatives. So the formula u tilde of 0 t equal to f of t plus t times f prime of t suggests that something worse may happen when f is not a differentiable function. To find out what may happen let us consider the Cauchy data f of r given by this formula f of r equal to 1 if r is less than or equal to 1 0 if r is bigger than 1. In other words this Cauchy data f takes the value 1 on the closed unit ball with center at 0 and outside the closed unit ball it is 0. The Cauchy data f of r is a smooth function everywhere except when r equal to 1 at which the function is discontinuous. Since f is discontinuous at r equal to 1 we expect trouble for u tilde of r comma 1 for r near 0 this is because f prime of t appears in the expression for u tilde of 0 t and f prime of 1 is not meaningful because function itself is discontinuous. Indeed u tilde of 0 1 let us compute it is limit r goes to 0 of u of r 1. We do not know whether limit exists or not let us compute. So what is u of r 1 I use the formula on the previous slides this is u of r comma 1. Now let us simplify this expression after the limit. So I take r common I get f of r minus 1 plus f of r plus 1. I have written r minus 1 as 1 minus r because f is an even function. So f of r minus 1 equal to f of 1 minus r. Now what is remaining is minus f of r minus 1 which once again I write as minus f of 1 minus r and plus f of r plus 1 by 2 r. I have not done anything I have just rearranged the terms in this expression to be this. Now there is a r here there is a r here I can cancel these two r. So I will separate this into two terms. This limit can be computed as limit of this plus limit of this provided these two limits exist. Now we see what is the limit of this and what is the limit of this separately. The first term is just 1 by 2 because when I am coming to r from the right side of 0 1 plus r is always bigger than 1 and f is 0 if the argument is bigger than 1. So this term is not there what I have only is this term and that term f of 1 minus r 1 minus r is always less than 1 and hence this is always 1. So 1 by 2. So in fact this quantity does not depend on r by the nature of the definition of the function f that we are considering. Then we have once again f of 1 plus r is 0. So I drop that term and I take this minus sign here limit r goes to 0 plus of f of 1 minus r by 2 r. f of 1 minus r is 1. So I have 1. Now I know this 1 by 2 r has no limit which is a real number but we will be speaking this is going to infinity and half minus infinity is like minus infinity. So there is no limit here. So after observing this limit equal to this minus this tells you that this limit exists if and only if this limit exists and we now just check this limit which is actually this limit that does not exist. So u tilde of 0 1 is not only not meaningful but also u tilde of r comma 1 goes to minus infinity as r goes to 0. Thus u is unbounded near the point x t equal to 0 1 x equal to 0 t equal to 1. Thus not only u of x 1 is undefined at x equal to 0 but also u of x 1 is unbounded as x approaches the origin. The singularity in f which was initially confined to a 2 dimensional surface which is a unit sphere gets concentrated at a point which is now the origin as time goes to 1 that is at the time t equal to 1. This is called focusing of singularities and one also says that caustics have formula. This is in complete contrast with the nature of propagation of singularities in one space dimension. In one space dimension the solution was as good or as bad as the Cauchy data. Now let us look at decay of solutions. Assume that the Cauchy data phi and psi have compact support. So throughout our discussion on decay of solutions we assume that the Cauchy data is having compact support. Otherwise we do not expect such results. Solution to Cauchy problem in 1d is given by the D'Alembert formula which is given by this formula u of xt equal to phi of x minus ct plus phi of x plus ct by 2 plus 1 by 2c integral x minus c2 to x plus ct psi sts. Even for a fixed x in R a large time behavior of solutions, solutions given by this formula is dominated by the term involving psi. We saw this in one of the tutorials where we saw the point wise u of xt converges to a certain constant in that example. Phi never played a role there. So due to this let us assume phi equal to 0. Now recall from lecture 4.6 and 4.5 the Poisson Kirchhoff formulae for the Cauchy problem in 2d and 3d which are given by this formula. Here the terms involving phi and psi behave similarly. Due to this we assume that phi is identically equal to 0. If phi is non-zero the estimates get modified by addition of new terms featuring phi and its gradient. The guaranteed decay rate of the solution will not change. In this lecture we are going to study decay properties of solutions to Cauchy problems for homogeneous wave equation across all the 3 dimensions 1, 2 and 3. We assume phi is 0 and Cauchy data has compact support. So we are going to assume Cauchy data has compact support. No decay for d equal to 1 we already saw this in example 2. We do not expect decay of solutions to one dimensional wave equation unless psi satisfies the integral over r equal to 0. In fact 0 to 1 in example 2 because we assume psi is supported in 0, 1. That is the content of example 2. So theorem for decay for d equal to 3. Let psi be a compactly supported function with support in the ball of radius r with center origin. Then for t positive we have the following 2 estimates. If you notice here this is a uniform estimate because this will go to 0 as t goes to infinity. As t goes to infinity RHS goes to 0. So this goes to 0 and the way it goes to 0 does not depend on x at all because this is valid for all x. The right hand side does not depend on x. That is why this is called a uniform decay estimate. The only difference between these 2 estimates is that here we have supremum of psi, here we have integral of normed right psi. So whenever these 2 things are meaningful we are assuming it has compact support so both are meaningful. So we have these estimates. Of course we do know that u of xt is actually 0 after some time for every fixed x. But this is a uniform decay estimates. So recall from lecture 5.5 that there were no sharp signal propagation for d equal to 2 but there is a decay of solution. This is what we mentioned. Even though the solution u of x naught comma t does not become 0 eventually that means there is a time after which u of x naught t is 0 for d equal to 3. That does not happen here. Something becoming 0 eventually means after some time it is equal to 0. That is the meaning of saying eventually 0. That does not happen. Nevertheless u of x naught comma t goes to 0 as t goes to infinity. In this lecture we will state 2 results on decay without proof. Of course even for d equal to 3 we have not given proof. For each fixed x naught a decay estimate on u of x naught comma t. That is one decay estimate. It is called point wise decay estimate because x naught is fixed and we are talking about the decay of u of x naught t. Second one is uniform decay estimate on u of xt. No proofs as the discussion is intended only for an exposure to decay properties as d varies. So decay at a fixed x that is the point wise decay. As before psi will be compactly supported function. Now in a disk of radius r centered at the origin in R2. Then for each fixed x there is a constant k which appears in this inequality that depends on x. Such that for all t this estimate holds. So as t goes to infinity right hand side goes to 0 and therefore left hand side also. But this estimate depends on k and k depends on x. That is why it is called decay at a fixed x or point wise decay estimate. Uniform decay estimate same assumption as before on psi. Then there exists a constant under t 0 such that whenever t is bigger than t naught this estimate happens. If you are interested as t goes to infinity it is okay. This goes to 0, t goes to infinity means t will become bigger than t naught after some time. So estimate is valid. So this goes to 0 therefore this goes to 0. So uniform decay. But notice it is a root t here in the point wise decay it was 1 k by t. Now it has become root t because of the uniform decay. So you are having a stronger statement here. So you have a weaker estimate that is what is true in general. So let us summarize propagation of singularities for d equal to 1 the nature of singularities in the solution did not change from that of Cauchy data at different times. For d equal to 3 the nature of singularities in the solution could be drastically different from that of Cauchy data. This is to be expected even from the Poisson Kirchhoff formula where even if phi is c 3 and psi is c 2 eventually the u you get is only c 2. Therefore if you fix a time t and look at u of x comma t naught for example this is not c 3 it is only c 2. So there is a loss of derivatives there. So therefore that is the reason behind this. Similar results are expected for d equal to 2. For d equal to 3 we observed that nice are weak singularities like discontinuities in the Cauchy data propagate and result in stronger singularities in the solution which was found to be unbounded near the point x t equal to 0 1 in that example that we saw. Decay properties of solutions of wave equation for d equal to 1, 2, 3 we have compared. D equal to 1 no decay, d equal to 3 there is a uniform decay of 1 by t type and for d equal to 2 we had a point wise decay and a uniform decay one was like 1 by t, one was like 1 by root t. Thank you.