 In the last lecture, we have started discussing method of characteristics for solving Cauchy problem for Quasi-Linear equations. In this lecture, we complete the proof of existence of solutions to Cauchy problem for Quasi-Linear equations using method of characteristics. So, outline of today's lecture is first we recall the assumptions and notations which are made about Cauchy-Linear equations. And then we describe method of characteristics. A quick recap of lecture 2.6 where we have completed this step 1 in the method of characteristics which is passing characteristics through points of gamma. Today, we start elaborating on step 2 where we say what is the finding a candidate solution and then show that that is indeed a solution. And then a small remark on the proof of existence and uniqueness theorem that we are going to prove at the end of the step 3. So, assumptions and notations in the context of Quasi-Linear partial differential equations, first order equations. This has been revisited many times 2.1, 2.5 and as last time also in lecture 2.6. So, Quasi-Linear equation we denote by ql it stands for this equation ux plus buy equal to c, abc are functions c1 functions on omega 3, omega 3 is an open subset of r3 such that a square plus b square is not equal to 0 at every point of omega 3. And the projection of omega 3 to x-ray plane is denoted by omega 2. Quasi problem given f space curve that is a curve in r3 which is given parametrically in this form where the parameter is running in interval i and fgh are c1 functions such that notation gamma 2 important, gamma 2 is a projection of gamma 2 x-ray plane that satisfies this condition. Gamma 2 is actually x equal to fs, y equal to gs as s varies in i. We want that f prime and g prime do not vanish at the same time at every point on gamma 2. Then we want to find a solution that means a function defined on some domain such that u on gamma 2 takes the value h which is a prescribed datum curve. Of course, we do not expect the solution domain to contain entire gamma 2 but a small portion of gamma 2 is also allowed. Let us look at the method of characteristics from which was discussed in the last lecture. The inspiration for method of characteristics is the following theorem which said give me a surface of the form z equal to uxy then saying that this in this surface is an integral surface is same as saying that this surface is union of characteristic curves corresponding to quasi linear equations. Since we want to construct an integral surface in other words wants to find a solution or a surface corresponding to a solution we will start with this idea of constructing some quantity some object which is union of characteristic curves. That is the idea behind this method of characteristics. So, step 1 is through every point of gamma you pass a characteristic curve. So, gamma is here we want to find a surface which contains this gamma. So, what we do is we pass through this characteristic curves like that and you see this is going to give a surface. So, characteristic this given datum curve is described by S and the characteristic curves for each fixed this point this characteristic passing through that is defined by T in R. So, we have a description of some object geometrical object in R 3 which is parameterized by 2 parameters S, T. So, as S and T vary we hope you will get a surface that was the idea. So, that is why we start with this step 1 where we pass characteristic curves like this. Then we say that the surface that we are going to generate would be of the following type where the third coordinate is going to be a function of the first 2 coordinates. That is what we hope and that we will achieve in the step 2. We will define a function u, u is a function of 2 variables x and y using inverse function theorem. And then we show that u is indeed the solution. Step 1 what we did is finding integral surface containing the datum curve or a piece of datum curve means we need to weave a surface as I have just shown you in the picture and if you weave that surface through passing curves through points of gamma and curves are characteristic curves then that theorem gives us a hope that the surface we get is going to be an integral surface. Here this is the datum curve these are characteristic curves this is the same picture which I have just drawn. So, for this what we do take a point on gamma any point on gamma looks like FSGSHS and then solve this system of characteristic ODE with initial conditions FSGS and HS it means that when you look at this gamma you take at this point FSGSHS the point looks like this this is 3 tuple and then you pass a curve through that. That will be denoted by XTS, YTS, ZTS this is what we get at the end of this step at t equal to 0 it is going through this point that is how we are solving we are giving the initial conditions to this system of characteristic ODE. Now by Cauchy-Lipschitz Picard's theorem it has a unique solution let it be denoted by like this X equal to XTS, Y equal to YTS, Z equal to ZTS. Where is it defined? S is in I and t belongs to JS which is an interval of course the interval contains 0 note the dependence of S on JS that is why it is written JS but we have seen a lemma on reparameterization which says we may take JS equal to R. So, now let us proceed to the remaining steps in this method. Second step is finding a candidate solution how do we get that? Of course we have to take help of this what we have been achieved so far. So, this is this mathematical object is described by two parameters and this lives in R3 so we hope it is a surface that is the idea. As I pointed out earlier S represents a parameter running on the datum curve gamma and t in R it is parameterizing characteristic curve passing through this point on gamma which is visited when t equal to 0. Of course gamma already lies on 1 by design initial conditions. The parameterized surface given by 1 I have put it in inverted commas here because it is not clear it is a parameterized object is clear surface means we expect something that is why I have put here till the confirmation comes. It is expected to be an integral surface because we were motivated by the theorem and therefore we hope that we get an integral surface. But this is described using parameters t and S now we need to write it as z equal to u of xy. So, what does this mean? It means that we would like to express the parametric surface given by 1 in this form z equal to u xy for some function u we need to find some function that means we need to find a domain on which this function is defined on these holes. In other words write the third coordinate as function of the first two coordinates using one that is the only thing that we have in our hands namely the family of characteristic curves passing through points of gamma. So, the strategy for finding u is this express t and S as functions of x and y we have x and y given in terms of t and S. Now, suppose I write t and S in terms of x and y go ahead and imagine this is the formula we write t equal to capital T of xy and S as capital S of xy. Now, we go and substitute in this relation z equal to z T S and define a function this way z of t equal to capital T xy S equal to capital S xy. So, we got some function of xy call it u xy this should work because z itself was motivated like that. The equation for z was derived after using that it is going to be the value of solution along the character along the base characteristic of x T S by T S refer to lecture 2.6. So, therefore it should work but there can be very technical difficulties we will see what they are. So, express t and S as functions of x and y like that recall that x and y given by x T S and y T S that together represent base characteristic curves that is this. Now, this suggests applying inverse function theorem to this function then you will get t and S in terms of x and y. So, if you want to apply inverse function theorem you have to see what is the setup of the inverse function theorem it involves a function it involves a C 1 function and certain Jacobian condition. So, we need to check all those hypotheses then only we can apply a theorem. So, therefore answers to the following questions will be needed what is the domain of this function is a function continuously differentiable is a Jacobian condition satisfied these three conditions. Usually one ignores questions 1 and 2 and just concentrate on third one at length. In fact, the third is not a condition it is a Jacobian condition satisfied is never checked that is put as part of the assumption and then one proves the theorem but still 1 and 2 are still important. So, you ask this question why you will understand later on. So, answer I will not give you think about think about this question. Okay, what is the domain of this function as we observed for each fixed S T varies in J S. So, it is going to be subset of R cross I for each fixed S T varies in J S. Okay, that is all we can describe it is not even clear what the set is is an open set etc. But anyway we will do away with this kind of answering these kind of difficult questions. So, it complicates further analysis. However, we may assume without loss of generality that A is equal to J cross I where J is a sub interval of R. I is where S is running the parameter S is running and I J is where the T is running. Okay, we take J equal to R this is what we said thanks to our lemma and reparameterization. So, therefore we may assume that without loss of generality that A is equal to J cross I where J is a sub interval of R. In fact, we can assume J is equal to R due to this lemma that I just quoted reparameterization of characteristic curves or by restricting to a small piece near gamma that is also okay. So, you have the point P at that point restrict to a small piece then it is true that one that J can be chosen independent of S. This idea comes when one discusses differentiable dependence of or continuous dependence of solutions on the initial data. So, see the ODE book by Wolfgang for this book is a good book we will be referring once more at least in this lecture okay, done. So, domain is R cross I fine or R cross some sub interval of I if you follow Wolfgang approach if you follow our approach it is R cross I. Now, next question is it C1 function if you look at this X and Y are solutions of certain ODE therefore with respect to T they will be C1 what about with respect to S? What about together? Answer is yes it is C1 function if partial derivative with respect to T and with respect to S both are continuous functions then it turns out that it is differentiable with respect to the two variables together as we know existence of partial derivatives of functions of several variables does not mean that the function is differentiable forget about that function need not even continuous at a point but all directional derivatives exists. But if partial derivatives are continuous then the function turns out to be differentiable. So, and it will be C1 therefore we are interested in XT, XS whether they are continuous functions XT as I said it is ABC they are nice functions. So, there is no problem what about with respect to S? Where is S? S is appearing in obtaining XTS and YTS through the initial conditions we solved X of 0S and Y of 0S as FS and GS and we assume F and GS C1 functions of the variable S that is the reason why this function is C1 of J cross I. It comes from differentiable dependence of solutions to initial value problems on the parameters in the theory of ODE's for instance see the book by Wolfgang on ordinary differential equations. Now let us implement step 2. Step 2 and question 3 we have to answer is the Jacobian condition satisfied by this function? We are interested in the invertibility of this function. So, we have to compute the Jacobian M-function theorem says compute the Jacobian that should be nonzero Jacobian is this. Now we have to compute XT at TS, YT at TS, XS and YS at the point TS. Now XT at TS is A it is known, YT at TS is also known, XS at TS is not known what we actually know is XS at 0 comma S that is the reason why we stick T equal to 0 now. Since we do not know the function explicitly we cannot compute JTS for an arbitrary point TS in J cross I which you can read it as R cross I. However using characteristic ODE and initial conditions because initial conditions for characteristic ODE are given a T equal to 0 therefore there something can be done. We can compute J 0S and that is given by this XT, YT are solving ODE's. So, from there you get AB X of 0 is 0S is FS therefore derivative of that with respect to S is F dash of S similarly YS is G dash of S. So, we want this to be nonzero. So, if you assume that J of 0 comma S0 is nonzero at a point S0 then we can apply the inverse function theorem I am written in the brackets local because there is something called something else called global inverse function theorem but the usual inverse function theorem is a local inverse function theorem thereby step 2 will be successfully implemented. This assumption on Jacobian is called the transversality condition. The quasi linear equation QL and the initial curve or the datum curve gamma are set to satisfy transversality condition. So, the transversality condition is two people together satisfy that that is quasi linear equation and initial curve. At a point S in I if the two curves which are listed here the first curve is a base characteristic curve. So, at point S right. So, look at the point FSGSHS that is going to be on the datum curve through that you find the characteristic curve project to R2 XY plane that will be the base characteristic curve. Thus base characteristic curve and gamma 2 which is the projection of the datum curve gamma these two curves intersect of course they intersect at the point FSGS non tangentially. That means the tangent lines do not coincide. If the transversality condition is satisfied at a point then it will continue to hold in an interval containing S0 because the transversality condition is non-zeroness of certain determinant. And whatever is appearing inside the determinant are continuous functions of S. Therefore, at some point non-zero it will continue to be non-zero near by that point that is the reason. In this case what would happen is imagine this is our gamma 2 that base characteristic curve will come like that. It will actually cut base characteristic curve will not be like this. For example, it will not be like that because here they share the same tangent. So, this is not allowed okay then transversality condition is not satisfied because if you see in the determinant what are the two things which are coming there the two columns. Xt yt is the tangent of the base characteristic curve and this is a tangent of the gamma 2 and asking that these non-zero means they are not parallel. And they are passing through this point FSGS 0, S sorry they are passing through the point FSGS yes and we are asking that the tangent line is not the same that is why it is the transversality condition. So, here it is transversal here it is not transversal. Now, if such a thing happens if you look at nearby also the curves all going to be like that right they will always be cutting and you get a surface that is a consequence. So, analytically the transversality condition translates to this being non-zero this determinant. It rules out the possibility of base characteristic curves touching the projected datum curve namely gamma 2 that we have illustrated in the picture touching means shearing the tangent line and coinciding of course. Then also you have touching for a long time this is maybe at a point but this is if you if you intersect along a piece of it it means that they share tangent throughout that piece because curve is same. So, such things are ruled out. Please spend some time on understanding this geometrically because it will be useful when we are trying to give examples or solve or guess what would happen in certain situations for quasi linear equations some examples it will help. So, this justifies the use of the word transversal it means to cut it cuts. So, assume that Jacobian is not 0 at the point s 0 even if Jacobian is not 0 for all s in i j of 0 s is not 0 for all s in i it is not even useful because the only way you apply implicit inverse function theorem is at a point s equal to s 0. So, I am assuming at a point it is non-zero of course s equal to s 0 means the point on gamma will be f s 0 g s 0 h s 0. By inverse function theorem there exists an open subset of j cross i containing this point 0, s 0 and an open set on the other side omega 2 containing the point x of 0 s 0 y of 0 s 0 because our function is that which is actually f s 0 g s 0 and a continuously differential function from f to e because this should be the inverse of the function phi. So, psi circle phi will be identity on e and phi circle psi will be identity on f these are the identity functions okay i e means now we have a picture see this is a thing phi we have e to f. So, originally phi is defined on some other set which is j cross i or r cross i or j cross some i dash and then we assume that at 0 comma s not Jacobian condition is satisfied non-zeroness of the Jacobian. Therefore, you can find an open set e which contains the point 0 comma s not and an open set f which contains the image of phi under that whatever is the map phi we have it maps t s to x t s y t s. So, x of 0 s not y of 0 s not which is f s not s not and a mapping psi such that these compositions you go from e to f come back from f to e or go from f to e first and then come back to f both will be identity functions this where we are using t s coordinates we are using x y coordinates the variables names. So, therefore, so psi of x y you denote it as t x y comma s x y and this equality is hold because they are inverses of each other that will give you t equal to t x y s equal to s x y. As I told you earlier z t s was designed to be the value of the solution at this point x t s y t s refer to lecture 2.6. Therefore, this motivates us the definition of a candidate solution by this you defined on f to r given by u x y equal to z of t x y s x y. As the function is a composition of 2 c 1 functions this is c 1 inside and z itself is c 1 function therefore the composition will be c 1 function therefore u is a c 1 function. Step 2 is successfully completed we have defined what is a candidate solution. So, in lecture 2.8 that is the next lecture we will discuss the geometric meaning of transversality condition and what happens when this condition is not satisfied what will happen when it is satisfied we are going to prove a result today and what happens if it is not satisfied will be analyzed in the next lecture. Can we still hope to solve Cauchy problem we have to see answers will be in the next lecture. Now, let us proceed to step 3. What is step 3? Candidate solution which we got in step 2 is actually solution by definition of u we have this formula and z t s equal to u x t s y t s we also have this. Now, what is that we want to show u solves PDE right. So, we have to compute u x u y etc. But that may not be useful let us see differentiating the last equation that is this equation with respect to t we get of course we use chain rule and we get this. Now, why did we not choose the first equation you could have chosen that differentiate with respect to x with respect to y go back and substitute in the equation and check whether it is satisfied or not this is usually the situation all the time whenever we have change of variables and PDEs are there we need to differentiate some equation but there are always two equations for that only one of them would be useful in the sense it may be a quicker way of getting the solution not wrong other one is not wrong but you may not be able to achieve anything with that but this one will definitely cure you will learn this as many times as you make mistake of differentiating this u that is true. Now, it is going to work differentiate we got this formula now x t s y t s and z t s are solution to Kara ODE. Therefore, here points will come out to be too long just to make the notation short I am using x t s y t s z t s as p t s. Now, z t is c at x t s y t s z t s but now it is p t s u x is as it is x t is a at x t s y t s z t s now it is a of p t s similarly the other term we have this. So, in terms of x y enough the last equation reads as c of x y u x y equal to u x of x y and a of x y u x y u y of b x y x y this means u is a solution to the Quasarine equation Cauchy data is already taken care we already observed that Cauchy data will be satisfied the datum curve will be on the integral surface. So, step 3 is also successfully completed after the three steps we have obtained an integral surface defined by a function u from f to r of course here we have used inverse function theorem that means existential to start with and s contains this point p 0 s contains a piece of the datum curve near p 0 in fact that much part of gamma for which f s g s belongs to this open set f if you could apply if there is a global inverse theorem if you could apply that then we would have got a solution defined still on a subset of omega 2 but that would have contained the entire datum curve gamma that integral surface would contain entire datum curve gamma the solution may still not be defined on whole of omega 2 this local global we had be very very careful what we have been dealing with is we are looking at a point on the datum curve and showing solution a surface exist nearby that which is integral surface that means we are discussing something like local with respect to the datum curve and if you say global inverse function theorem it must be giving global with respect to the datum curve but not with respect to the domain more on this we will see in a future lecture local there are two different concepts local with respect to datum curve local with respect to the domain omega 2. So, theorem is here hypothesis standard hypothesis on QL I am not going to recall here and the Cauchy data assume that the transversality condition holds at a point s 0 and denote p 0 f s 0 g s 0 h s 0 same thing let us call x 0 y 0 z 0 for convenience conclusions Cauchy problem has a solution defined on a neighborhood of the point x 0 y 0 and let d 1 d 2 be open neighborhoods of the point x 0 y 0 let u 1 u 2 be defined on d 1 d 2 respectively be solutions to Cauchy problem and then u 1 will be identically equal to u 2 on some subset of d 1 intersection d 2 of course it contains the point x 0 y 0 that is solution to Cauchy problem is locally unique near the datum curve gamma. So, let us prove conclusion one that is precisely steps 1 2 3 that gives the proof of conclusion one that is about existence now let us do the uniqueness let s 1 and s 2 be a as given one is defined on d 1 other one is defined on d 2 and take this point p 0 which is in gamma now p 0 is there and gamma it is there in s 1 as well as s 2 and as a consequence p 0 is there and s 1 p 0 is there and s 1 therefore a part of what is s 1 it is an integral surface therefore a part of gamma will be there on s 1 intersection s 2 it will be there on s 1 another part will be in s 2 there will be a common portion which will be there on both s 1 s 2 that is there. Then we have seen a corollary in lecture 2.6 in the last lecture it said some part of the each of the characteristic curves of q l passing through points of gamma dash will also be there on s 1 and also s 2 of course the corollary was with respect to one surface but here with respect to two or two surfaces yes you can see the corollary yeah this characteristic curve will be there on both s 1 and s 2. Due to transversality condition that is why step 2 is very very important step 2 the T s and x y the maps phi and psi are inverses of each other that is very very important. Due to the transversality condition any point on a characteristic curve has the form x y u 1 of x y and also x y u 2 of x y whenever you are in this common area f intersection d 1 intersection d 2 f you remember is the domain of definition for psi this proves that s 1 coincides with s 2 in a neighborhood of p 0. Hence u 1 is identically equal to u 2 on some neighborhood of x 0 y 0 this proves uniqueness now quick remark on the proof of existence and uniqueness theorem in fact it is going to be another proof of step 3 step 3 was very simple we had to simply compute the derivative substitute we see that the candidate solution is indeed a solution we can give a geometric proof of that surface s was described using this parametric equations right x t s y t s z t s fine the curves s equal to constant what are they they are characteristic curves this is gamma fix s on this this is x t s so if s is fixed it is a characteristic curve characteristic curve means what is the tangential direction first of all it is a curve therefore x t y t z t is a trans is the tangential direction so x t y t z t this is tangent but it is characteristic curve therefore the direction is actually a a b c remember this okay and characteristic curves lies on the surface s for s we know something which is normal okay characteristic curves lie on s x t y t is also a tangential direction to s and x t y t z t is nothing but characteristic direction we have observed what is the normal normal is a b c right no u x u y minus 1 yeah u x u y minus 1 that is a normal z equal to u x y if it is your surface normal is along this direction u x u y and minus 1 okay normal and what is the tangential direction one of them is a b c that has to be 0 and this is nothing but the equation therefore the surface s is an integral surface so any proof of the uniqueness assumption in the theorem is incomplete if the transfer surrogate assumption is not used because uniqueness fails with the failure of transversality condition when transversality condition fails there are examples where uniqueness fails there are examples where existence fails anything can happen so we will discuss that later so if you are proving uniqueness you must be using transversality condition is satisfied so people may not write this much explicitly in books but they will say that step two which is what we have said right they make use of step two without explicitly mentioning to you so you should not get confused the summary is that we consider the Cauchy problem for QL we proved the existence of a local solution assuming transversality condition holds nothing no more assumptions via method of characteristics we proved local uniqueness of solutions and in step two we noted a few possible reasons for existence of only a local solution which will be seen in next lectures and why you may not expect global global I already mentioned there are at least two notions of local solutions and hence global solutions so in the next lecture we will take up as promised what is transversality condition geometrically and what happens when it fails what are the possibilities thank you