 Welcome to this lecture on the Cauchy problem for wave equation in 3 space dimensions. We are going to deduce a formula for solution to the Cauchy problem. This solution formulas are known as Kirchhoff formulae or Poisson formulae. So instead of giving credit to one of them, we give to both of them and call them or refer to these formulas as Poisson-Kirchhoff formulae. So the outline is very simple. We start recall certain things that we did in lecture 4.4 which is the basics about the spherical means and a couple of results related to that and then we go on to derive Poisson-Kirchhoff formulae. So the key ideas in solving the Cauchy problem in lecture 4.4 we found an equivalent Cauchy problem in 2 independent variables rho and t for any number of space variables wave equation. The equivalent Cauchy problem was obtained by following method of spherical means. When d equal to 3 some nice things happen after change of dependent variable in the new Cauchy problem we obtain a Cauchy problem for wave equation in one space dimension and Dallambert formulae is readily available here. A solution to the Cauchy problem for wave equation in 3 space dimensions is then retrieved thanks to LOSM of lecture 4.4 LOSM is lemma on spherical means. So let us recall from lecture 4.4 definition of spherical means for a continuous function we define spherical mean of g as a function of x comma rho denoted by mg of x comma rho which is integrate g over the sphere S x rho multiply with 1 by mod S x rho. S x mod S x rho is the surface measure of S x rho. So we have defined the spherical means by this formula and we have obtained an alternate formula and this formula has an advantage over the first formula because the domain of integration does not depend on x or on rho. So lemma on spherical means the hypothesis is that if you have a continuous function and define the spherical means f mg then mg can be extended to rd cross r such that for each fixed x in rd the mg as a function of rho is an even function. So let k belongs to n if g is a ck function on rd then so is this function x rho mapping to mg of x rho it is a function defined in rd cross r note by one conclusion one we have already extended the function mg of x rho for rho belonging to r and the function g itself can be recovered from the spherical means. So if you know the spherical means for radii rho which are going to 0 then mg of x rho limit as rho goes to 0 will give you g of x this will happen for every x in rd for g in c2 of rd the Darbu formula is Laplacian of mg with respect to x variable is this dou 2 by dou rho square plus d minus 1 by rho dou by dou rho acting on mg. So this is the so called radial Laplacian. So equivalent Cauchy problem to introduce that we need to define the spherical means of u. So let u be a c2 function rd cross r be a solution to the Cauchy problem for d dimensional wave equation then define m u the spherical mean of u as usual for the same formula for the spherical means. Now this m u satisfies a different Cauchy problem that is what we are going to say here. So equivalent Cauchy problems if u is a solution to the Cauchy problem for d dimensional wave equation then the m u solves the following Cauchy problem. This is the radial Laplacian here c square and this is second derivative with respect to t of m u and these are the initial conditions for m u and dou by dou t of m u. So this is still not the wave equation we need to change the dependent variable m u to something else then the right hand side will look like the wave equation. That means right hand side looks like dou 2 by dou rho square of that new quantity. So this is the Cauchy problem for 3 dimensional wave equation. Now when d is equal to 3 the d minus 1 is 2. So this is the equivalent Cauchy problem that we have derived in the last lecture that is lecture 4.4. Now we are going to change the dependent variable m u this is not the usual wave equation in one space dimension. So we are going to change it to rho m u that we will call as l l of rho t equal to rho m u. This l will satisfy a one dimensional wave equation let us do the computation. So dou 2 by dou rho square of l rho t is dou 2 by dou rho square of rho m u because l equal to rho m u. So keep one dou by dou rho out take one by dou by dou rho inside when you differentiate the inside quantity with respect to rho you get this plus this with this is rho times dou m u by dou rho which is here plus m u. Now differentiate once again with respect to rho you get end up with this. So this is actually this rho this operator dou 2 by dou rho square plus 2 by rho dou by dou rho acting on m u. So using this equation we get this equation this in terms of l. So what we have is l satisfies dou 2 by dou rho square equal to 1 by c square dou 2 by dou t square which is nothing but the wave equation in one space dimension. And let us write down the Cauchy data for l. So l of rho 0 will be this and dou l by dou t of rho 0 will be rho m sin. So here we have a one dimensional wave equation and the corresponding Cauchy data therefore using the Alambert formula we can write the solution. Now the question is is the Alambert formula going to give you a classical solution. To get that what do we need is this function should be c2 because l of rho 0 should be c2 and dou l by dou t of rho 0 should be c1. So therefore the question is is this c2? This is c2 if and only if m phi is c2 with respect to rho right this is just rho multiplication by rho is a very good thing. So this quantity is c2 if and only if m phi is c2 this quantity is c1 if and only if m psi is c1. So are these conditions met? Answer is yes. By l by sm2 it said if g is ck, mg is ck. Therefore what we need is phi is c2 therefore m phi will be c2 and hence rho into m phi will be c2 which is l rho 0. And we need psi to be c1 so that dou l by dou t is c1 at rho 0 is c1. Now we ask another question why are we assuming more regularity and phi comma psi we will see this soon. This is not good enough. What is not good enough? This assumption is not good enough for us to deduce a solution to the Cauchy problem for 3 space dimensions from the D'Alembert formula that we get. This hypothesis is good enough to apply D'Alembert formula to the equation for l and get the expression for l of rho comma t stops there we will point out later. So using the D'Alembert formula this is the l rho mu is the l. So l equal to rho minus ct m phi plus rho plus ct m phi plus 1 by 2c into the integral term because this is psi and this is the phi of this problem. So the above equation gives us m mu you want therefore you divide everything with a rho so we have this and this okay. So we have an expression for m mu now I want u therefore LOSM3 tells me I can get u from m mu but I need to pass to the limit as rho goes to 0 therefore we need to pass to the limit on the RHS the quantities on the RHS as rho goes to 0. Okay so let us do that this is the expression for m mu. So RHS has two terms for ease of presentation we handle them separately and we will be using the first conclusion of LOSM many times what is that spherical means or even function with respect to rho. Okay the first term is this we want to pass to the limit in this term if you look at the numerator numerator goes to 0 as rho goes to 0 because this goes to minus ct m phi of x comma minus ct plus ct m phi of x comma ct and m phi is even with respect to the radius variable therefore m phi of minus ct is m m phi of ct therefore it gets cancelled and you get 0 and of course limit of 2 rho is also 0. So therefore we are in position to apply L hospital rule therefore the limit rho goes to 0 of this is precisely the limit of the derivative of the numerator the denominator is of course just 2 so that is why it is 1 by 2 1 by 2. So the quantity inside the brackets is the derivative of this with respect to rho. So it is very clear how do we get these terms differentiate this with respect to rho then you get m phi of x comma rho minus ct. Similarly, this with respect to rho here this term you get m phi of x comma rho plus ct product rule. So you have to differentiate this with respect to rho that will give you this rho minus ct times derivative of this with respect to rho Similarly here rho plus Ct times derivative of this with respect to rho. Now the first term tends to m phi of Ct as rho goes to 0 because as rho goes to 0 this goes to m phi of x comma minus Ct plus m phi of x comma Ct but both are same because of LOSM that means I have 2 times and then I have a 1 by 2 here therefore I get m phi. So it is the easiest term to pass to limit. Now let us look at the second term and we just write down the formula for m phi. m phi formula is 1 by omega d whether here it is 1 by omega 3 into integral on norm nu equal to 1 of phi of x plus rho minus Ct nu. Similarly, this term also we can write like this. Now omega 3 is 4 pi and we can put that and the derivative term the dou by dou rho is outside the integral. So let us take it inside and differentiate we get this because when dou by dou rho goes here I get grad phi and derivative with respect to rho will give me nu so that is why nu here. This omega 3 is 4 pi therefore this became 8 pi. Similarly, the second term also we can write after pushing the derivative inside the integral. So this is what we have. Now we want to pass to the limit in this term as rho goes to 0. We obtain this quantity just straight forward from here and the second term will give you this. So I take t by 8 pi common I have this expression. Now I pulled dou by dou t outside of the 2 integrals and the one which is inside both the terms are equal therefore that is 2 times that. So when it comes out you get t by 4 pi into dou by dou t of this. So we have finished obtaining the limit of the first term. So as rho goes to 0 limit of the first term on RHS of this equation is this. The first term gave us m phi the second term gave us this. Now we need to now pass to the limit in the second term here that is much simpler. This we can club these 2 terms like this. If you expand this you will get this. So this is a more compact notation compact form of this formula. So we identified the limit of the first term let us do the identify the limit of the second term now. So this is the limit we are interested in. The integral term goes to 0 why as rho goes to 0 this is integral minus ct to ct and m psi is an even function with respect to the variable s but there is a s which is multiplying therefore that will be an odd function therefore the integral is 0 when rho is 0 and of course denominator also goes to 0 once again L hospital rule that tells us the limit is 1 by 2c times derivative of this which by fundamental theorem of calculus is this and now we have to take limit of this as rho goes to 0 and that is very simple. Now once again m psi is an even function so it further gets simplified 2t into m psi. So now we have obtained all the necessary limits and we are in a position to write the formula for u. So uxt is equal to this this is the limit of the first term this is the limit of the second term. So the above formula is known as Poisson Kirchhoff formula some books say these are Poisson formula some books say they are Kirchhoff formula we use both the names. This formula that we derived let us call f1 we are going to derive another formula from this just by expanding this integrals you get some other expression f2. Now these integrals are on norm nu equal to 1 so we can change back them to integrals on the sphere s of x comma ct then we get 2 more formulas f1 will become some other formula f2 will become some other formula. So these are the other 2 formulas f3 and f4. So we have f1, f2, f3, f4 formulae which all give solution of the wave equation Cauchy problem for the wave equation we have to be slightly careful here when we say that. Of course there is let us first discuss the advantages of f1, f2 over f3, f4 here the exactly the same reason the domain of integration does not depend on x t here also in the both these formulae whereas in the next 2 formulae they depend. Now we want to check that u is a solution to the Cauchy problem first you have to check whether you use a c2 function. Now if you look at this formula does not look like it is going to be a c2 function suppose phi is c2 then grad phi is c1 if psi is c1 this is c1 so the entire quantity may at most look like c1 that is the reason why we put additional hypothesis on phi and psi. If you recall in one space dimension wave equation the Cauchy data was assumed to be c2 and c1 the initial displacement and initial velocity we have assumed one is in c2 other than c1 now we have to jack up the smoothness otherwise this function u will not be a classical solution perhaps we can say that this will be a weak solution once we define what is the notion of weak solution etc which is beyond the scope of this course therefore let us put additional hypothesis that is what we are going to do. So Poisson Kirchhoff formula they are derived under these assumptions phi is c2 psi is c1 that is enough because how did we get Poisson Kirchhoff formula starting from the Dalambert formula and for the Dalambert formula we needed just this assumption phi is c2 psi c1 so that the Dalambert formula gives a classical solution to the equivalent Cauchy problem after that we just pass to the limits we do not require any further regularity so these are good enough to derive a formula for a possible candidate we should not say solution possible candidate to the Cauchy problem which we have derived but it will not be twice differentiable under the these hypothesis that we have just discussed so to guarantee that u is twice differentiable we assume that the Cauchy data satisfies phi is c3 and psi is c2 that is the reason. So the following result says the Poisson Kirchhoff formula represent a classical solution to the Cauchy problem proof is left as an exercise phi c3 psi c2 then this formula is a solution in fact checking this formula will necessarily go through converting these integrals where the domain of integration does not depend on x and t I guess that is what it is or else you should know a formula straight away what is the derivative of this that must be derived once so in any case it is left as exercise so even though all the four formulae represent solutions the first two are more convenient for verification using f3 or f4 one has necessarily go through one of the first two formulae in some form or the other the secret no secret really we have revealed this many times the first two formulae f1 or f2 the variables t and x do not appear in the domains of integration unlike f3 or f4. So let us summarize we have derived Poisson Kirchhoff formulae for the solution to Cauchy problem for wave equation in 3D for defining the solution u only the following smoothness is required phi is c2 and psi is c1 only this much is required however to guarantee that u is a c2 function we need more smoothness on the Cauchy data which is phi is c3 and psi is c2. So even though Cauchy data is smoother the solution u at a later time becomes less smooth okay so this was not the case in one dimension so something happens in 3D and we will discuss these kind of issues later. Thank you.