 In the last lecture, we considered Cauchy problems for Burgers equation where the Cauchy data or the initial data was not smooth or even when the initial data was smooth we found that in one of the examples that solution is only piecewise smooth. So such functions cannot be solutions in the usual sense which we described as classical solutions. So we asked the question is there a framework under which we can admit such functions also as solutions to Burgers equation and in general for a first order partial differential equation. So we look into that in this lecture. So do not be afraid by the word conservation laws we are not going to study too much about it, it is only in the context of Burgers equation that we are going to discuss. Burgers equation can be written in what is called a conservative form we will come to that. So this is a brief recall from the last lecture. We considered 4 initial value problems for Burgers equation. We came across 2 kinds of difficulties first one was a solution could not be determined in some region of the upper half plane because there were no base characteristics and a solution becomes multi-valued due to too many base characteristics entering a particular region that is another reason why we could not define what is the solution there was some ambiguity. So the notion of solution we considered so far prior to the Burgers equation is often called classical solution. As a consequence initial value problems do not admit global solutions we already understood that. If we relax the notion of solution then initial value problems may admit global solutions that possibility we will get. A relaxed notion of solution is possible for Burgers equation due to its conservative form which is given by ut plus u square by 2 this bracket x stands for dou by dou x differentiation with respect to x. So assuming u is smooth when you do expand this by chain rule what you get is ut plus u ux equal to 0. We have used u y when we solved it by characteristics method and but I told you that y has the interpretation of time so therefore it is ut this is fine. The relaxed notion is very natural for equations in conservative form. We are going to introduce one relaxed notion. Of course, conservation laws are not new to us we have seen them already in traffic modeling. So this notion is called generalized notion that we are talking about is called a relaxed notion is called weak solutions to conservation laws. So there is a guideline for relaxing the notion of a solution. Even for a conservation law we can demand that it should be a nice function differentiable function so that ut plus u square by 2 differentiation with respect to x equal to 0 is actually ut plus ux equal to 0. But we want to admit you know functions which are not so much smooth as solutions therefore we would like to define a new concept of a relaxed solution and there should be some guidelines. What are those? Three requirements are there. Any notion of relaxed solution we may call it weak solution must have the following three properties. What are they? Any smooth solution should also be a weak solution. This is to be expected it should be there right. Otherwise you are defining some new solution. Earlier we said we want to relax because some equations may not have the smooth solutions or a classical solutions. But if they do have classical solution we would like that the relaxed notion also it admits as a solution. Therefore smooth solution should also be a weak solution. This is usually the guiding factor in defining any notion of weak solution. We will soon see how that is going to be done. And any weak solution if it is smooth then it should be classical solution. This is a second requirement of course. So we prove that the notion of a relaxed solution which we are going to give motivated by 1. So we are going to define what is called a relaxed solution or a weak solution concept. And then we show that any smooth solution is a weak solution. And then we also show that any weak solution which is smooth should is actually a classical solution. We will show that. These are the two and then the third one is the most important thing and we are not going to address that. Any reasonable problem should have a weak solution. Otherwise what is the use of notion of a solution when you cannot show that such a solution exists when the problem is reasonable right. I am not expanding what is reasonable but this is what one has to remember. These are guidelines. Discussion of this requirement for the notion of weak solution we are not going to do that is beyond the scope of the course. So let us look at the initial value problem for a conservation law. Burgers equation in the conservative form is a special example of this where f is u, f of u is u, g of u is u square by 2. So assume that this equation has a classical solution that means a differentiable solution so that this you can expand f dash u into u y, g dash u into u x equal to 0 by chain rule and u x 0 equal to h x. This u of x 0 makes sense and is equal to a pre-assigned h x. So here f, g and h are a priori given to you. Smooth functions. So let us see how we arrive at a notion of weak solution. First thing is to take a function phi which is compactly supported in this domain r cross closed interval 0 infinity closed at 0 and c infinity that is it is differentiable any number of times. C infinity function with compact support and the domain is r cross 0 infinity 0 closed. And first thing is your equation you have to multiply with phi. This equation is multiplied with phi and integrate. So this equation is simply this equation multiplied with phi the two terms are separated and integrate on your domain r cross 0 infinity. So what have you achieved? Nothing we could know we have not yet relaxed that you can be a lesser smooth function for non-differential function etc. Therefore first thing the moment we see a derivative here and a derivative here first thing is idea is to shift this derivative to c infinity function that we have. So therefore we have to do integration by parts in this integral with respect to y in this integral with respect to x. So with that we get from here we get this integral and from here when we do we get this integral and in this integral there is a one boundary term which is here because we are taking phi which are compactly supported close 0 infinity not open 0 infinity. If it is open 0 infinity this term will not be there because phi x 0 will be 0 in that case but we are taking with this that is because we want to account for the initial condition that is what is going to come here. So this is what you get at the end of integration by parts. So what we observe just now is that any classical solution satisfies this integral equation and for every phi which is c 0 infinity r cross 0 infinity and this equation if you see it is meaningful even for you which are not c 1 for example u is only continuous f of u makes sense right f of h of x of course makes sense g of u makes sense and these integrals are on infinite domains but phi is having compact support. So essentially the integrals are on a bounded set therefore when you integrate continuous functions and bounded sets it is integrable these are well defined integrals okay. So therefore this is meaningful even for you which are not c 1 I just use the word u is continuous of course you do not need even u to be continuous what all you need is this should make sense. A notion of weak solution gets defined once we mention what kind of functions u we would like to allow them as solutions. So we have to decide which u you are going to allow as solutions for your problem then a notion gets defined you will ask that this integral equation should be satisfied for all phi in this space and you should lie in some space that you have to identify you already decide. So you fix the class of function that you are interested in or you like with the only condition that this equation is satisfied for every phi you will get a notion of weak solution. So let us give one notion of weak solution here assume that h is L infinity of r this lobe should not be there because L infinity functions are bounded functions or if I put L infinity lobe of r it just means that it is bounded function on every compact set that is good enough. So L infinity lobe is good we can keep this L infinity just means that Lebesgue measurable functions which are bounded essentially bounded functions whenever there is a lobe it means that on compact sets some property holds L infinity lobe means it is in L infinity of every compact subset of r. Now u in L infinity lobe of this set see now we just want bounded measurable functions as solutions not even bounded everywhere bounded on every compact set that is good enough for this notion in particular non-differentiable functions all of them will come under this they satisfy this condition. So u in L infinity lobe is said to be a weak solution of the initial value problem which is here if for every phi c is infinity this integral equation is satisfied. This was first of all derived from this equation assuming that you use a smooth solution multiplied with a phi coming from this space and then we found this is satisfied. Now we forget all that and we say as long as this is satisfied I am happy and now I put some conditions so that this makes sense and we are demanding it should be equal to 0 one such class is L infinity lobe r cross 0 infinity ok. The previous definition if we restrict phi to compactly supported functions infinitely times infinitely many times differentiable r cross open 0 infinity then the boundary integral the integral and r will vanish it will be 0 as I pointed out phi of x 0 is 0. So we are left with only these 2 times then u is called a weak solution to the conservation law it not Cauchy problem for the conservation law but weak solution to the conservation law that means we are worried only about the equation and not the initial conditions. The notion of a weak solution to the initial value problem is also referred to as its weak formulation that integral equation which we saw on the previous slide is often called a weak formulation of the conservation law. Now this is what we have finished the guiding principle part 1 all smooth function smooth solution should satisfy the new formulation that is how we have derived. Now we will go to step 2 where we are going to show if you have a weak solution and it is smooth it must be classical solution that is what we are going to establish now. So for this we need to assume slightly one extra condition on f of course for burgers equation f of u is u of course that is a 1 1 function. So let f be a 1 1 function let u be a smooth function that is c 1 because it is a first order PDE c 1 is required and here I put continuity up to 0 that means u of x 0 makes sense. If you have continuity up to this 0 close 0 that means that u of x 0 makes sense. Suppose this is a weak solution to the initial value problem then u is a classical solution of the initial value problem that is the conclusion. So here we are assuming u is a weak solution and smooth then we are going to say it is a classical solution of the initial value problem. So since u is a weak solution and also smooth what does it mean? I can go back the strongness and the weakness what is the connection integration by parts. So I need to do integration reverse integration by parts from the weak formulation. So this is this is the meaning of what we have here it is a weak solution to the conservation law. Now if you do integration by parts you will get back this okay this is the boundary term that is going to come. So what happens if I take a phi which is c 0 infinity open 0 infinity this term will not be there we will get this. Since the integrand is continuous right integrand is continuous and phi c 0 infinity is arbitrary fundamental M i in calculus of variation essentially it means if you integrate against c 0 infinity function a certain function and you always get 0 then that function must be 0. So it is essentially like this just imagine something like this we have phi psi equal to 0 for all phi in c 0 infinity functions omega that would imply that psi is 0 under various assumptions psi it is true definitely when psi is continuous it is true. So therefore use a solution to the conservation law. Since use a solution to the conservation law this equation now just becomes the last term because these two together are 0 so we have this. Now once again phi is arbitrary therefore this must be equal to this and if the function f is 1 1 inside thing must be equal to inside thing that is the idea. So this is true for every phi in c 0 infinity r cross close 0 infinity now any if you notice this is only function of x here this is only function of x it is not like y there is no just function of x integral is an r. So we can get any smooth function c 0 infinity function of r through this phi any psi in c 0 infinity r looks like phi of x 0 for some phi here for example phi of x y equal to psi x into chi y where chi is psi c 0 infinity r that is given to you and chi is c 0 infinity of close 0 infinity with chi identically equal to 1 in some interval 0 1. So when you put y equal to 0 chi of 0 is 1 therefore you get psi of x so that is simply this. Now once again you apply this is a continuous function integrate against any c 0 infinity of r function is 0 then this function must be 0 that result I am loosely calling it as fundamental lemma and calculus of variations. So using that f is a 1 1 function we get u x 0 equal to h x otherwise to start with you get f of u x 0 equal to f of h x since f is 1 1 you can take away the f and you get u x 0 equal to h x. Now we are going to look at another question which piecewise smooth functions are not weak solutions the following theorem will be helpful in deciding that suppose you have set d subset of r cross 0 infinity and suppose you have a curve that divides d into two parts. So generally one writes this kind of picture this is d and you have a curve gamma and it cuts you that this into two pieces d 1 and d 2 and suppose u 1 is c 1 and continuous up to the boundary d 1 closure means up to boundary that means u 1 is here it is smooth similarly u 2 is here. So u i u 1 is c 1 of d 1 and also continuous up to the boundary so that I want to talk about the values of u 1 on gamma similarly I want to talk about u 2 values of u 2 on gamma. So I require c 1 of d 2 and continuous up to d 2 closure so that u 2 on gamma is also meaningful. So two conditions generally people write as an intersection so this and this and u 2 is c 1 in the second domain and continuous up to its closure. Define a function u on d now like this u 1 in d 1 u 2 in d 2 okay and gamma we are not defined. Now let this bracket u denote the jump in the values of u across gamma that means we had this this is d we had d 1 here d 2 here. Now u 1 on gamma makes sense at points of gamma similarly u 2 on gamma makes sense so we can look at the jump u 2 at a point p on gamma minus u 1 at the point p this is the jump jump in u at the point p you could also define u 1 minus u 2 at point p I have used u 2 minus u 1 both are same you have to have just a consistent way of defining it. So this is the definition at any point x y which is on gamma you define the jump in u as u 2 minus u 1 at that point so we have defined what is the jump at all points on gamma. Now similarly you define f u jump in f u and jump in g u okay jump in f u at a point p is f of u 2 at p minus f of u 1 at p that should be the definition jump in f of u at a point p is on d 2 the value is going to be u 2 right f of u 2 at p minus f of u 1 at p that is a jump similarly g jump in g is also defined. Then the following statements are equivalent saying that u is a weak solution to the conservation law is same as saying that u 1 and u 2 are classical solutions in the of the conservation law in both the domains d 1 and d 2 u 1 in d 1 and u 2 in d 2 and along the curve gamma certain condition holds which is f u n y plus g u n x equal to 0 what is n x n y it denotes the unit outward normal to gamma that means we had like that we had a gamma take a point so normal direction this is tangential direction this is a normal direction you can take either this or this any one of them you choose does not matter that is what we are saying so let us choose for example this does not matter because the equation anyway is equal to 0 if you replace n x n y with minus n x minus n y it is the same okay. So further if the curve is described by x equal to xi y this is an interesting assumption it is given by x equal to xi y what does it mean the curve looks like xi of y comma y okay that means it is a graph with respect to the y variable then the rank on given your condition takes this form d j by d y equal to jump in g by jump in f what is the importance of a curve given by x equal to xi y see y is a time variable right y is time variable so so it is like xi of t right x equal to xi of t it is giving you some points in x which are the values of somebody at t so it is telling you the location of somebody as time evolves so when y is time variable the curve tracks the location of points if the initial data is an x axis it will track the location of the points from x axis tracking points of discontinuity or nondifferentiability of the initial data where u x 0 is h x h x is the initial condition so these kind of things can be tracked by this curve that is why we always assume x equal to xi y so proof of 1 implies 2 in the definition of the weak solution to the conservation law using phi I am allowed to use any phi which is c 0 infinity of d right now I use c 0 infinity of d 1 and d 2 we get the first part that is it is a smooth solution it is a solution because of smoothness we do integration by parts now let us derive the Rankine-Hugonio condition along points of gamma so take a phi which is c 0 infinity of d now I can not do di if I take di I am ignoring or the other di right if I take d 1 c 0 infinity of d 1 I am forgetting d 2 no because the jumps are across gamma gamma is there for both d 1 and d 2 right it is abounded so therefore you take a c 0 infinity of d and the weak formulation of the conservation law reduces to this and since d is a disjoint union you write this integral as integral over d 1 plus integral over d 2 okay now I think we have to do integration by parts so if n x n y denote unit outward normal to d 1 then minus n x minus n y denotes unit outward normal to d 2 because this unit outward normal is what appears in integration by parts so performing integration by parts we get these terms dou by dou y has gone to on and on d 1 u is u 1 u is u 1 on d 1 similarly here u is u 2 on d 2 u is u 2 on d 2 that is why you get f u 1 y from here g u 1 x from here and this is the boundary term coming from this integral one with respect to x one with respect to y so this is the other one or you can also treat this as a Green's theorem okay this is okay I will leave it to you we call integration by parts by various names depending on the context fine u 1 and u 2 are classical solutions to conservation law we already proved therefore some things will be 0 okay the domain integrals will go off what remains is a gamma integral okay this is what you have now the notation n x n y is fixed okay n x n y is outward normal to d 1 then we have used that minus n x minus n y is outward unit normal to d 2 while doing integration by parts so finally we end up with this gamma integral now phi is arbitrary therefore integrand must be 0 and which will give you Rankine-Hugonio condition if the curve is given by x equal to xi y we compute the normal normal direction is 1 comma minus d xi by dy therefore Rankine-Hugonio condition becomes this equation d xi by dy jump in g by jump in f 2 implies 1 is very simple so that is left as an exercise a few remarks on the last theorem we already started with this remark when we stated theorem the curve gamma described by x equal to xi y for some function xi this means that gamma is the graph of a function of y the curve gamma is a curve of discontinuity for weak solution in burgers equation the variable y has the interpretation of time variable and thus y going to xi y tracks the location of discontinuity from its initial location as the time evolves so theorem is useful in rejecting the candidature of a piecewise smooth function for a weak solution to IVP how you simply show that Rankine-Hugonio condition is not satisfied along the discontinuity curve and therefore it cannot be a weak solution so it is useful in rejecting the candidature theorem analyzed solutions with jumps across one curve why not consider functions with jumps across more than one curve very natural question it is important while dealing with initial profiles which have more than one point of non-smoothness theorem is applicable along every curve of discontinuity as long as the functions considered are piecewise smooth functions caution is there exists multiple conservative forms multiple okay for burgers equation we have written one right u y plus u square by 2 of x equal to 0 and these are so many as many as natural numbers so we have here at least at least so infinitely many are there now which one will you consider you have to decide so note that r h condition also depends on k because it is jump in g by jump in f that is the equation for d z by d y now g varies this is the g new g right depends on k similarly this is f depends on k so r h condition depends on k so what does that mean if you take two different case and take u and v which are weak solutions corresponding to the different case then the vice versa may not be true so therefore in applications you have to realize which one is the most correct or useful conservative form of your equation also so weak solutions are not unique because we have relaxed the notion of solution now lot more functions we do not expect uniqueness now that is what is demonstrated by these examples now we are going to use the following conservative form throughout this lecture from now onwards u y plus u square by 2 x equal to 0 let us write down what is the rank in Egonio condition it will turn out to be u 1 at point of gamma plus u 2 at the point of gamma by 2 that means the average of u 1 and u 2 so for example this is the one d 1 d 2 u 1 here u 2 here take a point take the value of u 1 and take the value of u 2 and take the average that should be the d xi by dy the slope with respect to y here okay this is the example 1 that we considered in the last lecture with initial data minus 1 and 1 minus 1 up to x less than 0 1 for x greater than or equal to 0 it has many weak solutions few of them are being given here for example here we have 2 curves of discontinuity one is the line x equal to minus y another one is the line x equal to y now let us see whether our r h condition is satisfied here what is the average of u 1 and u I mean this is d 1 this is d 2 here so u 1 and u 2 average x by y on this line is minus 1 because the line itself is x equal to minus y therefore x by y is minus 1 therefore what is the average of minus 1 and minus 1 it is minus 1 so x equal to xi y that is the curve we are considering and we have to look at d xi by dy so in this example d xi by dy is minus 1 and which is also equal to the average therefore r h condition is satisfied across this line now let us discuss across this line across this line one side the value is 1 other side is x by y but on this line x by y is 1 therefore average is 1 and that is same as x equal to xi y derivative of xi is 1 so Rankine-Hugonew condition is also satisfied across this line therefore this is a weak solution let us look at this here the curve of discontinuity is x equal to xi of y equal to 0 so therefore d xi by dy of course is 0 and that is same as the average of minus 1 and 1 minus 1 plus 1 by 2 is 0 so this is also a weak solution let us look at this in this case d xi by dy is this quantity minus 1 plus a by 2 minus 1 plus a by 2 and the average is also that minus 1 minus a divided by 2 which is precisely this here d xi by dy is 1 plus a by 2 and that is precisely the average 1 plus a by 2 here xi y is 0 d xi by dy is 0 for this curve d xi by dy is 0 and that is also the average minus a plus a by 2 is 0 so here also R H condition is satisfied across all the 3 discontinuity curves let us look at the example 2 which is 1 and minus 1 this also has many weak solutions is one of them here d xi by dy is 0 because this equation x equal to xi of y equal to 0 therefore d xi by dy is 0 and that is the average of 1 and minus 1 so R H condition is satisfied here it is 1 minus a by 2 and that is precisely d xi by dy this is the xi of y so d xi by dy here minus 1 plus a by 2 and that is d xi by dy so R H condition is satisfied across this line also what about this line also because minus a plus a by 2 is 0 and that is xi dash of y xi of y is 0 therefore xi dash is also 0 so R H condition is satisfied across all the 3 lines. Let us look at the example 3 look at this here of course there are many pieces this is one line of discontinuity this is another line this is another line let us look at the top line this line x equal to y plus 1 by 2 what is xi dash for this line half y by 2 right xi of y is y by 2 plus 1 by 2 derivative will be 1 by 2 and that is precisely the average 1 plus 0 by 2 so across this line R H condition is satisfied. Let us check whether across this line R H condition is satisfied across this line 1 minus x by 1 minus y is nothing but 1 because on this line x is equal to y so therefore 1 minus x by 1 minus x u is 1 from this side u is 1 from this side therefore average is 1 and that is also xi dash of y xi of y is 1 therefore xi of y is y therefore xi dash of y is 1 therefore R H condition is satisfied here also. Now let us look at this line x equal to 1 on this line x equal to 1 this u is 0 because it is 1 minus x by 1 minus y x equal to 1 means it is 0 by 1 minus y so u is 0 this side u is 0 this side so average is 0. Now what about xi of y is derivative xi of y here is 1 and its derivative is 0 therefore R H condition is satisfied across all the 3 line segments. Now the concept of entropy solutions we are not going to study too much about it but just to get some awareness. So we saw that there are too many weak solutions is there a special one somehow we are longing for a unique solution. So we saw that classical solutions may not exist for all T positive even for smooth initial data for Burgers equation we have seen. We relaxed the notion of solution to a weak solution we did that but we got too many solutions too many weak solutions defined for all x in R and for all T positive we just saw. Thus our notion of weak solution is too much relaxation can we find a good weak solution? This is where the notion of an entropy solution appears entropy solution is people call it is a physically relevant solution in examples of physical importance they have introduced this entropy concept of entropy solution and that is indeed a physically relevant solution and even for a mathematical problems which are very far from applications any entropy condition people call it a physically relevant solution there is no physics behind that. Entropy solution is what is a desirable solution because it fixes some solution uniquely even in a mathematical problem. So entropy solution is at most one that is how the notion gets developed. So further discussion on entropy solutions may be found in the books by Bresson and also the book of smaller that I have earlier referred to in the last class. So with this we conclude the discussion of first order PDEs. So what did we see in this lecture? How a notion of weak solution evolves it was demonstrated minimum requirements that any reasonable notion of weak solution must satisfy they were discussed. A notion of weak solution was developed for burgers equation using a conservative form a characterization for curves of discontinuity for piecewise smooth weak solutions was obtained that was called Rankine-Hugonio condition. Using the above condition we identified some functions which are piecewise smooth solutions to burgers equation and in fact we have all the example that we have seen they are all weak solutions only we found them. So weak solutions are not unique in general and thus one needs to impose more conditions on the notion of solution so that the problem admits only one solution after this imposition of new rules. And this lecture is intended mainly to say that whatever initial conditions and solutions that we considered in 2.15 lecture of course they are not classical solutions but still they have some meaning and they can be given meaning in the sense of weak solutions and further continuation is entropy solutions. Thank you.