 Welcome, in this lecture we are going to introduce notions of generalized solutions to wave equation. The outline is as follows, first we mention that the notion of a classical solution is inadequate and thus there is a need to generalize the notion of a solution and then in step 2 we show how to arrive at a generalized notion of a solution and then we demonstrate some generalized solutions to wave equation in 1D and in higher dimensions. In this lecture, we discussed the inadequacy of D'Alembert's formula, Poisson-Kirchhoff formula for a classical solution to the Cauchy problem for wave equation for describing physically relevant situations. We present the standard procedure to arrive at notions of generalized solutions also known as weak solutions for the purposes of this lecture to Cauchy problem for wave equation. Generalized notions of solutions to IBVPs may also be defined similarly. So recall the Cauchy problem for wave equation, here the data is given phi, psi and f and we are required to solve non-homogeneous wave equation D'Alembert's u equal to f and ux0 equal to phi x, utx0 equal to psi x for x in Rd. Recall from lecture 4.7 where we have derived the solution as given by this formula in 1D. This formula represents a classical solution if phi is C2, psi is C1 and assumption on f is that fx is continuous apart from f itself being continuous. So when the string is plucked at a point x0, what does that mean is this? Suppose this is a point x0, so string is pulled up so as a consequence how this string looks is like this is lying up to here and then let us say like that, like that and like that. So at this point x0 it is raised, this is a graph of u of x0. This is a situation clearly such a function cannot be differentiable at this point x0. The initial profile that is a graph of phi of x looks like a triangle as we have seen. In this physically relevant example the function phi is not differentiable at x0. Therefore this formula does not make sense as a classical solution because the function will not be u of xt will not be C2 because phi is not C2 it is not even differentiable. So therefore this formula is inadequate in this scenario. So in such a case where the Cauchy data is not smooth this formula let us still call it Dalambert formula even if we have a right hand side the source terms here right hand side in the wave equation. So this let us still call it by the name Dalambert formula. So this does not give a classical solution to the Cauchy problem. Worse Cauchy problem may not have classical solutions. Not only that this formula does not is not a classical solution it may not even have a classical solution. Thus there is a clear need to change the concept of a solution. So can we require some last ground? See this is the formula we have derived in 1D right for the solution of Cauchy problem. It is a classical solution if these conditions are satisfied by the data phi is C2 psi is C1 f should be continuous on this domain and fx should also be continuous on this domain. The formula itself makes sense for much bigger classes of functions because what all you need for this phi can be any function because there is no differentiability requirement on this. It can be any function this makes sense. Here I have a psi I need to integrate therefore psi should be integrable. But this is for every x and t it should be integrable on the interval x minus ct and x plus ct. Psi need not be integrable on whole of r it is enough it is integrable on every interval of this form or more generally any interval of the type a comma b closed interval a comma b. Similarly f, f has to be integrable on certain as we already observed this domain is nothing but a triangle. This integral integration is done on a domain which is a triangle which is called characteristic triangle. So we need f to be integrable on that. So let us see some assumptions on phi psi f which guarantee that the right hand side is meaningful therefore it defines a function. The question then is is it a solution or in which sense is a solution etc. For example phi psi are continuous functions as I told you there can be any functions f is continuous and hence it will be integrable on any triangle that you take. Psi is continuous therefore this integral is meaningful for every x and t that is what I have written here phi and psi are L1 loke. L1 loke simply means that this function is integrable on every compact set or equivalently on every bounded set. So similarly f is assumed to be L1 loke locally integrable this is a notation for that. Please note that we are using this notation L1 loke just to mean these are locally integrable functions it means functions are integrable on every compact set. Please do not confuse this with the Lebesgue function spaces of L1 L1 loke because these kind of things the point wise evaluation do not make sense if I say that this is the Lebesgue spaces but when we see the weak formulation their L1 loke it makes sense so there is no problem. So whenever we see point wise evaluations like this L1 loke stands simply for those functions which are integrable on every compact set and when we see in the weak formulations this stands for the usual Lebesgue L1 loke spaces. Now look at the formula in 2D. 2D formula is slightly more complicated than the formula in 1D because here phi comes along with the derivative also psi appears like before so it is not a problem but phi has a derivative. Now these are classical solution to the Cauchy problem where phi is C3 and psi is C2 and f should have this kind of smoothness properties. But the formula itself is meaningful for the following classes of functions when phi is C1 so these are continuous functions and you are integrating on closed disk so it must be fine integral will exist similarly psi continuous so integration is not a problem and f continues. We can further weaken the smoothness requirements on phi psi f by saying phi should be C1 that seems to be there seems to be no alternative because grad phi appears in the formula. Now we are looking for conditions on the data so that these integrals make sense. So phi should be C1 because grad phi is there and psi is L1 and f is L1. Of course this is not the exhaustive list of classes of functions there are much bigger class of functions for which this is meaningful and discussion of such things is beyond the scope of this course. Now solution in 3D same issue here because phi comes with a derivative. So the formula represents a classical solution to Cauchy problem if phi is C3 psi is C2 and f has the regularity or smoothness properties which is given here. Formula itself is meaningful when phi is C1 psi is continuous and f is continuous or more generally psi you can allow to be local integrable function and f also local integrable function. So from the last few slides it is apparent that each of these formula delambed formula Poisson Kirchhoff formula defines a function even when the Cauchy data and source term do not have enough smoothness to guarantee that the function is a classical solution. We have identified a few classes of functions to which such Cauchy data and source term may belong to in order that the expression for u of xt is meaningful or makes sense. This gives us a hope for the recovery of some last ground due to lesser smoothness of the data. Indeed there exist notions of weak or generalized solutions which admit such functions u as solutions to the Cauchy problem. We understood the need to admit functions which are less smooth as solutions because Cauchy data itself we have to admit which are lesser smooth. It may happen that solutions defined by this formulae delambed or Poisson Kirchhoff may give rise to a classical solution in a restricted xt domain. We have seen that such things happen in the case of burgers equations and conservation laws. We may use a formulae to study how the lack of smoothness in the Cauchy data propagates with time that is propagation of singularities. Lack of smoothness means it is called a singularity something is not smooth at some point that point is called a point of singularity. So, propagation of singularities in the Cauchy data with time. This we will discuss in lecture 5.7 arriving at a generalized notion of a solution how to do that. So, there are guidelines for relaxing the notion of a solution. We have seen this in lecture 2.16 where we have introduced in the context of burgers equation and conservation form of that. Three requirements on a relaxed solution. Any notion of a relaxed solution or weak solution or generalized solution whichever word you may use it must have the following properties. What are they? Any smooth solution should be a weak solution. This requirement is usually the guiding factor in defining any notion of a relaxed or generalized or weak solution. Any weak solution which is smooth enough should be a classical solution. Proving that the notion of a relaxed solution that we are going to introduce soon motivated by one the point one that is the requirement one has this property is left as an exercise because the ideas are very similar to what we did in lecture 2.16. Now the third requirement any reasonable problem should have a weak solution or a relaxed solution. Any notion of a solution is useless when reasonable problems do not admit solutions. Discussion of this requirement for the notion of weak solution that we are going to introduce soon is beyond the scope of this course. We will not discuss that. So, deriving a notion of generalized solution to Cauchy problem all the details will be presented in one space dimension. For higher dimensions obvious modifications are needed and we get a similar formulation we are going to show that formulation. So, let zeta be a C2 function on R cross R this 0 stands for compact support. That means support of zeta is a compact set. So, support is contained in a big enough ball equivalently. So, multiply the given equation the non-homogeneous wave equation with zeta and integrate with respect to xt over this domain R cross 0 infinity what we get is this that is simply integrating both sides after multiplying with zeta you get this equation. There is nothing much to do this side because you really do not know f to do anything. Here we can do something here we see that there are derivatives on u and we are not looking for smooth solutions to the wave equation therefore we would like to relax this requirement and that can be done by transferring these derivatives to zeta which is integration by parts. So, nothing much to do on RHS we would like to integrate by parts on the LHS. So, this is the LHS first term in the LHS because there is also a dou 2 dou x square term. So, this on integration by parts you get this one derivative shifts to zeta and a minus sign will come and there will be a boundary term with respect to this 0 infinity t 0 infinity. Because t equal to 0 is a lower limit you will get another minus sign minus zeta x 0 dou u by dou t x 0 upper limit will not contribute because zeta has compact support. So, this is what we have but what is dou u by dou t of x comma 0 that is ix. So, we have got so we have done one integration by parts. So, let us do once more transfer this dou by dou t to this one that will make this minus as plus and the transfer is done and similarly as you had this term coming out of the integration by parts we will have one more term coming from this. So, that is this and the other term which we had here is written here. But now what is u of 0 it is phi x. So, this is what we get by integrating by parts in the first term on the LHS. Now, let us take up the second term which involves zeta dou 2 by dou x square. Again integration by parts here there will be no boundary term because with respect to x zeta is supported once again compactly and we do not have this situation of 0 infinity in when we are doing with respect to x that is why there will be no boundary terms that is why 0. So, it is do once more integration by parts we get this no boundary terms because integration by parts when you do you get an integral on the domain and you get an integral on the boundary. So, this is what we have. So, this equation we are going to use the standard notation this we call D'Alembertian 1 right square 1. So, in that notation I am going to write so I have just interchanged the LHS and RHS because this is where we are going to write expression for this. So, expression for this that we derived using integration by parts that is why I am writing here. So, this is what we get in fact what we got by integration by parts is this term on the RHS equal to this term on the RHS. So, what we did is that any classical solution to the Cauchy problem satisfies this equation which we have derived. So, this is the requirement 1 in fact that was a guiding principle now we are going to define what is a weak solution based on what we have got here. So, the above equation is meaningful for you meaningful for you which is not necessarily C2 because nowhere there is a derivative on U U is only here right nowhere else zeta is a smooth function with compact support C2 function with compact support. So, this is a continuous function with compact support essentially you need to integrate on a compact set. So, if U is L1 locally integrable this is meaningful. So, a notion of weak solution gets defined once we mention what kind of function we would like to be solutions. So, in d space dimensions any classical solution satisfies this equation derivation is exactly same. We have shown the derivation when d equal to 1 but exactly the same steps will give us this. The last equation is meaningful for locally integrable functions not only U for phi psi f also. So, we are now in a position to define a notion of generalized or relaxed or weak solution. Let phi and psi be local integrable functions and f be local integrable function on this domain r decrease 0 infinity. Let U be also an L1 log function the function U is said to be a weak solution to the Cauchy problem for the wave equation if this equation is satisfied by every zeta which is C2 uncompactly supported in rd cross r. The requirement 1 is automatically satisfied every classical solution to Cauchy problem is a weak solution. The weak formation itself was derived based on this requirement. Proof of the requirement 2 as I told before is similar to the idea which we presented in lecture 2.16. So, if U is smooth then it can be shown that U is a classical solution by doing integration by parts in a weak formulation. So, in the definition of the weak solution we had an equation right valid for all zeta in C02 C20 functions that is called weak formulation of the Cauchy problem. So, U is smooth means this because this is what we need for a classical solution. If all those conditions are met by our weak solution then it is actually a classical solution and that is left as exercise. Important questions it was easier to define the notion of a weak solution it is always easy to define something that is not a difficult job. Do we know some of the weak solutions? Do the Dallambert formula Poisson-Kirchhoff formula represent weak solutions to the Cauchy problems? This is the question. As mentioned earlier we are not going to discuss the existence of weak solutions. We limit ourselves to checking whether the named formula quoted above namely the Dallambert formula or Poisson-Kirchhoff formula are weak solutions. Let us look at some of the generalized solutions to wave equation in 1D. Recall U of xt equal to f of x minus ct plus g of x plus ct we have obtained this as a general solution to the wave equation whenever f and g are c2 functions. Now we are going to ask the obvious question is this a weak solution when f and g are not c2 functions. So we are going to show that this formula f of x minus ct plus g of x plus ct is a weak solution even if f and g are not c2. We will show that f of x minus ct is a weak solution. Similarly we can show that g of x plus ct is a weak solution therefore it follows some of 2 weak solution is a weak solution because equation is linear and homogeneous. Here what do we require if you want to show f of x minus ct is a weak solution. Definition of a weak solution wants local integrable functions so we can as well take f2 may be local integrable function. Since we are checking interested in checking that f of x minus ct is a weak solution to the wave equation and thereby we are not worried about the Cauchy problem. So this weak formulation where we insisted that zeta is c2 0 of r plus r now it reduces to exactly same equation without these terms and zeta we can take to be open 0 comma infinity and automatically these terms are not there. So essentially f is not there homogeneous wave equation so f is 0 and phi and psi are not there because zeta is compactly supported in open 0 infinity. Therefore what we have is just this and I want u of x t the candidate I am proposing is f of x minus ct. So we want to show this equal to 0 we need to show this. So let zeta be compactly supported function c2 in defined on this domain r cross 0 infinity. We would like to do integration by parts here but unfortunately we cannot do that because we are assuming f is not c2 function therefore I cannot transfer all the derivatives to this by integration by parts. So we have to do something else in particular we are looking for f to be a weak solution therefore this will be a l 1 low function we want to show that this integral is 0. Therefore there is no way we can do integration by parts but here somebody helps us with namely change of variables. Let us set x minus ct equal to xi and x plus ct equal to eta then this integral becomes this integral this is f of xi and this wave operator dou 2 zeta by dou t square minus c square dou 2 zeta by dou x square becomes minus 4 c square w xi eta of xi eta. W is nothing but xi eta expressed in xi eta coordinates then dx dt become d xi d eta by 2 c and the integration domain is the is upper half plane is given by eta bigger than xi. Now we can do integration by parts in this formula with respect to eta because f is l 1 low but depends only on xi so there is no eta dependence. So once we do integration by parts here the domain integral will be 0 and only boundary contribution remains and the boundary of this domain is precisely eta equal to xi which is nothing but the x axis. So therefore we should know what this quantity is and this is equal to 0. Here w of xi eta is nothing but zeta of xi plus eta by 2 comma eta minus xi by 2 c. So we will differentiate this with respect to xi and evaluated eta equal to xi. As the support of zeta is contained in r cross 0 infinity w xi of xi xi is nothing but zeta x derivative of zeta with respect to x at the point xi comma 0 into derivative of this with respect to xi which is 1 by 2 minus 1 by 2 c zeta t at the point xi comma 0 and that is equal to 0 because the support of zeta is contained in the upper half plane and here evaluation is happening because of the t equal to 0 we are on x axis therefore this is 0. Let us look at the example 2 of a weak solution. Dalambert formula whenever meaningful is a weak solution to the Cauchy problem. However the proof is not as simple as the one we saw in example 1 in which a change of variable yields the desired result. We will handle this proof using approximation procedure. In lecture 4.11 an IBVP for wave equation was solved a formal series was proposed as a solution. It was shown to be a classical solution if phi was C4 which is unreasonable we saw the example of plug string is unreasonable. However if phi is C2 the series nevertheless converges uniformly and it defines a function and that function let us call it u of xt because the series converges this is meaningful so that will be only be a continuous function not more than that. So it cannot be a candidate for classical solution but it can be a candidate for weak solution. So the nth term may be written as forget about this are numbers look at this term that can be written as this using sin a cos b formula we get this. Now something interesting is happening here this is a function of x plus ct this is a function of x minus ct. So each of the terms here is sum of 2 terms which is a function of x plus ct another one is a function of x minus ct. So we can split this series into 2 series then what we will have one series will be a function of x minus ct one series will be a function of x plus ct and we whenever you see a function of x minus ct alone it is weak solution to the wave equation homogeneous wave equation similarly other one x plus ct so and therefore it is a weak solution this represents a weak solution to wave equation. Let us look at another example in lecture 5.1 the following equivalence was established for a c2 function what was that u is a solution to the homogeneous wave equation if and only if u satisfies the parallelogram identity on every characteristic parallelogram. Parallelogram identity is meaningful for any function it just involves values of the function at the 4 vertices of the characteristic parallelogram a relation between the values of the function. So if a function u satisfies the parallelogram identity on every characteristic parallelogram then is it a weak solution to the wave equation that is the question. What we have shown is that if u is c2 then it is a classical solution to the wave equation. Now the question is u is any function which satisfies parallelogram identity nothing more than that no extra hypothesis on u being c2 therefore we ask this question is it a weak solution of course to define the concepts of weak solution you have to start with the L1 log function. So let us assume u is L1 log function can I do this. The answer to the above question is yes. So any function u satisfying parallelogram identity on every characteristic parallelogram may be written as u of xt equal to f of x minus ct plus g of x plus ct for some functions f and g this proves that you use a weak solution to homogeneous wave equation. In view of the arguments we gave in example 1 because the sum of two functions one is a function of x minus ct alone another is a function of x plus ct alone but how do I get this f and g I am given a function u which satisfies the parallelogram identity on every characteristic parallelogram how do I get f and g how do I catch them. At this point I would suggest you to pause the video try to find f and then g using the requirement which is here this is a requirement before you resume the video it is fun okay let us see how to get these functions f and g in the characteristic coordinates it is always easy it is easy to find f g such that w xi eta equal to f xi plus g eta now exactly do the same exercise as before pause the video try to find f and g from this requirement first you get f and then you get g once you try for some time then even if you are not successful it does not matter but you have tried so now you can easily follow what is being done on the next slide I have also tried like this I was not successful I try to define a function f of xi I have defined then g of eta when I am using this equation directly that is involving w xi eta minus f xi okay that is not a function of eta alone because xi is also there then I also paused for some time thought about it and then I got the answer so please try yourself okay let us see now fix l and k in R because we have to use parallelogram identity right without using that any formula that you try to do will not work I am sure about that once you try you will also be sure okay so fix l and k any f and g as above should satisfy this right if they exist w of lk must be fl plus gk so define f of xi equal to w of xi k g of eta equal to this now you see there is no xi in this expression that is a trick which works so let us compute f xi plus g eta I have to show if this is equal to w of xi eta so this computation will give us this and that is equal to w xi eta because these 4 points are the vertices of a characteristic parallelogram but in the characteristic coordinates system it is a characteristic not parallelogram but a rectangle so these are the vertices and you know the parallelogram I will do holds therefore this all the combination of 3 terms is giving you w xi eta so we have shown what you want to show but in characteristic coordinate system writing the proof for x t coordinate system is left as an exercise to you if you recall in lecture 5.1 we have shown the equivalence of a function being a solution to the homogeneous wave equation and if it satisfying parallelogram identity on every characteristic parallelogram in that proof we have presented only in x t coordinate system but if you think of a proof in xi eta coordinates it is very easy as I have presented here in this context so what we have done there you have to imitate that here in x t coordinate system so that is a hint for solving this exercise so please go back and watch lecture 5.1 again carefully. Try the proof of that theorem in characteristic coordinate system and see how we have written the proof in x t coordinates now exactly you should be able to do that by translating these ideas into x t coordinate system. Now let us look at some generalized solution in higher dimensions so in d space dimensions any classical solution satisfies this that is how we have defined the notion of weak solution based on this in particular when the Cauchy data is smooth as required the function you would define by Dallambert formula or Poisson Kirchhoff formula are weak solutions okay if you are thinking of d which is greater than or equal to 2 we need not mention the word Dallambert formula we would like to show the same even if data is not that smooth as required for classical solution but good enough to make these named formula namely Poisson Kirchhoff for D equal to 2 and 3 and Dallambert formula for D equal to 1 they are meaningful and define a function. So the idea is to approximate the non smooth data f i psi in the Cauchy problem with smooth data f n phi n psi n what is the meaning of smooth data smooth data means it is that data which when you take the Cauchy problem it has a classical solution and is given by those named formulae let that solution be denoted by un un is a classical solution corresponding to this data once it is a classical solution it is also weak solution. So therefore this is satisfied now I have written this with n everywhere okay it will be very nice if you can pass to limit f n goes to f u n goes to u pi n goes to phi psi n goes to psi so that the corresponding integrals will go to the corresponding integral where there is no n that is a dream always if the convergence is good enough to allow passage to the limit in the above equation then we would get that u is a weak solution. So what is the requirement on the data f i psi should be such that the named formulae define a function in 1D define a function which is also l 1 look in 1D f i psi in l 1 look is enough as we already observed in 2D 3D the formula feature derivative of phi therefore we assume phi or phi is c 1 f psi can be l 1 look so requirement on the approximations there should be as smooth as required to guarantee that the named formulae give classical solutions the following convergence must be guaranteed also which is that f n goes to a phi n goes to phi psi n goes to psi and u n goes to u because you please note that u n you have a formula in terms of f n phi n and psi n so u n has a formula and that u n should go to some function u in l 1 look then we can pass to the limit in this integral and get that u is a weak solution. So remark on weak solutions in higher dimensions we would not further elaborate on the existence of smooth approximations as proposed our intention was to present the ideas that would convince us to deal with bad Cauchy data without worrying much we dealt with many examples involving discontinuous data and we promised to justify it later and we have fulfilled to an extent in that promise so constructing approximation as required needs more background in analysis and this is another reason for skipping the technical details but the idea we have presented that is good enough. So we had the requirement on the data f phi psi in 2d 3d the Poisson Kirchhoff formula featured derivative of phi so we assume phi is c 1 so what to do if phi is discontinuous in fact one of our examples in lecture 4.6 there we have solved problems in higher dimensions there we have used a discontinuous function for phi so what happens perhaps we need to go for a weaker notion of solution than the notion of weak solution introduced in this lecture let us summarize what we did in this lecture so during our discussion on wave equation we have derived in many a context formulae for solutions if data is smooth enough the formulae gave rise to classical solutions at times we need to work with non-smooth data in such cases there are questions on what actually the formulae stand for in all those contexts we have given an assurance that we can make sense of the formulae to yield solutions in a generalized sense this lecture is an attempt to introduce notion of generalized solution in the context of wave equation so this is at the expository level we have not presented really details in this lecture d equal to 1 d'Alembert formulae parallelogram identity series solution for IBVP they were analyzed and for their candidature for weak solutions if you recall wherever we use d'Alembert formulae with discontinuous initial conditions parallelogram identity and the series solution we promise that we do not have to worry there may be some new notion or generalized notion of a solution which will admit them as solutions go ahead and compute so this lecture was an attempt to justify those statements d equal to 2 3 we had partial success in interpreting Poisson Kirchhoff formulae as a weak solution in the sense that we needed to assume phi is c 1 we assume that that is why only partial success we could not handle phi which are discontinuous functions in such cases as I mentioned earlier one has to introduce notions of more weaker solutions than weak solution. Thank you.