 So, today we will take the first steps towards solving Cauchy problems for quasi-linear equations. So, in this lecture we are going to look at the relation between characteristic curves and integral surfaces. The outline of the today's lecture is, first we discuss an idea to solve Cauchy problems for linear and semi-linear equations which are clearly the simplest form of the quasi-linear equations, special forms but simplest. So, we try to solve the Cauchy problem for linear and semi-linear equations and we see whether we encounter any difficulties. If we encounter we see how to overcome them and when we generalize to quasi-linear what should be done will also be discussed. And then we look at the relation between characteristics and integral surfaces for quasi-linear equations. This we are bad to look at this as a consequence of our analysis for the 0.1. When we look at linear and semi-linear equations and try to solve the Cauchy problem naturally we are led to asking certain questions that will lead us to our understanding of this. It will lead us to enquire into the relation between characteristic curves and integral surfaces which we will do more generally for the quasi-linear equations. So, first we discuss an idea to solve Cauchy problem for linear and semi-linear equations. Let us recall from lecture 2.1 where we have introduced the notion of Cauchy problems and quasi-linear equations. So, this is the definition of the quasi-linear equation. It is an equation of the form AXYUUX plus BXYUUY equal to CXYU. Any reasonable analysis of this can be done at an elementary level. For that we need to assume that A, B, C are nice functions that is C1 functions. They are functions of 3 variables so they are defined in a domain omega 3 in R3. And we want that A and B do not vanish simultaneously that is A, B is not equal to 00 at every point of omega 3 so that we have some form of the PDE. If both A and B vanish then there is no PDE right LHS is 0 therefore we require that at least one of this A and B is non-zero at every point of omega 3. So, what is Cauchy problem? Here we are given a space curve gamma which is typically given in a parametric version where X component is given by FS, Y is GS, Z is HS as S varies in an interval in R and the functions FGH are C1 functions on the interval. This is given. So what is the Cauchy problem? It is to find the solution of the quasi-linear equation but before that we need to assume something more that is this projection gamma 2 of gamma of this space curve to R2 to the X-ray plane is what is called regular curve what it means is a prime, G prime is not 00 at each of the points of gamma 2 and we require that this Z equal to HS right that HS must be given as U of XY that is what we expect so U of FS GS that should be equal to HS. This is what we want to do and as discussed before we do not want S belonging to the entire interval I but we are happy if we can find the such a function with these properties where S belongs to a sub interval of I. That is a part of the gamma lies on the surface Z equal to UXY. Now what are characteristic curves for linear and semilinear equations they are the trajectories of the solutions of system of characteristic ODEs. In the linear case dx by dt is A dy by dt is B dz by dt is of this form CZ plus D. In the case of semilinear equation it is CXYZ basically ABC right AUX plus BUI equal to right hand side here also AUX plus BUI equal to the right hand side. So these are the characteristic system of ODEs their solutions and the images of the solution they are the characteristic curves. System of characteristic ODE take a pool into two sets of equations that is what we need to observe. If it is a quasi linear equation we will have A of XYU, B of XYU and C of XYU and because A and B do not depend on the third variable the first two equations if you look they feature only X and Y equation for X and Y can be is given in terms of right hand sides which are functions of XY only. So we do not require to solve this equation we can solve these two independently. Similarly in the semilinear equation X and Y can be solved for with without bothering about the equation for Z. So it decouples into two sets of equation a system of two equations for X and Y which do not involve the variable Z we saw that and a single equation for Z which can be solved provided you give me what is X and Y. Given a solution XT by T are the first two equations we can solve for Z. In other words base characteristic curves can be determined directly without the knowledge of characteristic curves. Remember characteristic curves are the traces of the triple XT, YT, ZT as T varies the trace of this is a characteristic curve and when you project it to XY plane you get XT, YT as T varies but here XT, YT can be determined independent of ZT therefore we say base characteristic curves are determined directly without the knowledge of characteristic curves. It is possible since the equations for X and Y do not involve the variable Z otherwise base characteristics are defined as the projections of the characteristic curves to omega 2. So an idea to solve Cauchy problem for L and SL we are dealing both L and SL together mainly because the part where the first order derivatives appear is the same AUX plus BUI okay. What is the step one? Fill up omega 2 with base characteristics passing through gamma 2. So what you do for this? Take a point X0, Y0 in omega 2 pass a base characteristic curve through that point. How do you do that? It can be done. You have to solve this initial value problem. This is the equation for base characteristic curves the first two equations dx by dt by dt equal to a and b respectively and you solve with this initial condition X0 is X0, Y0 is Y0. So such solutions exist because we are assuming a, b are C1 functions. So they are locally Lipschitz this is an initial value problem and it has a unique solution which defined on some interval containing 0 all that is there. Therefore there is a base characteristic through the any given point X0, Y0. Now the question is we are worried about gamma 2 that is where the data is given. So question is will it pass through a point of gamma 2 that is the question we ask and because that is where the data is given right we need to solve Cauchy problem. So that is why we ask this question it will be clear in the next slides why we are asking this question. On the other hand we can do one more thing take any point in gamma 2 why you are taking omega 2 take on gamma 2 okay and then pass a base characteristic curve through that yeah it can be done exactly same procedure you need to solve this initial value problem where X0, Y0 there is replaced by FSGS. Now the question is can we fill omega 2 with such curves by varying points on gamma 2. So these are the two different points of you. So questions from step one are the following right take any point in omega 2 pass a base characteristic curve through it will it pass through a point of gamma 2 one question. Second take any point on gamma 2 pass a base characteristic curve through that point question is can we fill omega 2 using such curves. Answers to the first question is may not pass through second question is may not be able to fill okay for both question we do not seem to have outright positive answers. Both are answers both answers are having caution terms may not may not but what we want to do we want to solve Cauchy problem. Therefore what is more relevant to us is gamma 2. Therefore we work with curves which are constructed in this two okay therefore that is the reason why we are going to look at curves which are defined using this point 2 here. So we take points on gamma 2 pass base characteristic curves and then try to find a solution that is the approach we will take because the initial data the data is given on gamma 2 x equal to FS y equal to GS that is gamma 2 z has to be HS is the datum curve. So with this decision we are giving importance to solving PDE near gamma 2 yeah because near gamma 2 why we say because we may not be able to fill up entire omega 2 by using the curves that we get in as an answer to step 2 okay we may not be able to fill that is the answer right. So we may not be able to fill but definitely we will cover gamma 2 so we should be happy with that to start with. So anything similar to ODE initial value problems here yes there is there are similarities there also when we solve dy by dx equal to f of x y and y x not equal to y not that is initial value problem the existence theorem says there is a solution near x equal to x not that is why it is called local existence theorem okay so it is similar to that. Now what is the step 2 solution is found along each base characteristic curve what is that we will explain. So let x equal to x TS and y equal to y TS denote the base characteristic curves passing through a point FS GS in gamma 2 so this is what we have obtained in step 1. Now I am writing this in quotes it means that it may not be very rigorous but this is a feeling okay when we prove theorems everything is a rigorous. So a solution u can be determined along each of the base characteristics found in step 1. So setting Z TS equal to u of x TS y TS I look at what is the equation this Z TS satisfies because what is this u of x TS y TS means I take a base characteristic curve x TS y TS okay for a fixed S, S is fixed here and then u of x TS y TS is there I can consider now that I call Z TS this is actually u along the curve x TS y TS that is what I am calling Z TS. So equation for Z TS is given by of course your differentiate dz by dt equal to u x into dx by dt plus uy into dy by dt but dx by dt and dy by dt they are solutions of base characteristic curves they will be A and B respectively. So what you have here is A u x plus B u y and that is equal to C if you use a solution to the PDE therefore this is equal to C of this is in the linear case C of x TS y TS into Z plus D of x TS y TS and if it is semilinear equation it is simply C of x TS y TS Z. So we got an equation a ODE for the solution along each base characteristic curve. So the equation for Z is a linear ODE for the linear PDE and it is a non-linear ODE for the semilinear PDE non-linear because you know the dependence of C on Z is not linear that is all it says. So for L the equation for Z was this I am recalling the equations it is a linear ODE with variable coefficients because the coefficients depend on T here also this is a non-homogeneous you can take this to the other side what you have is dz by dt minus C of x TS y TS into Z equal to D of x TS y TS. So you have a right hand set term and you have a variable coefficients here. Now Z TS can be determined for all T in J okay what is J? J is the interval on which you have solved x TS and y TS you have to remember that this equations for base characteristics were solved so that the initial condition was FSGS therefore this interval J actually depends on S I do not want to confuse you at this moment therefore I am not writing that dependence otherwise we have to write T belongs to J S we will see this in a forthcoming lecture I will highlight that point okay. So Z TS can be determined for all T in J such that Z 0 is HS why is that because linear ODE's have global solutions while non-linear ODE's have local solutions as a rule okay non-linear ODE's having global solutions is exception linear ODE's will always have global solution global solution means wherever your coefficients are defined in this case the coefficients are all defined whenever T belongs to J therefore there is a solution for T belongs to J. Now let us comment same about the semi-linear equation this is the equation this is a non-linear equation therefore Z TS may not be determined for all T in J it means that you may have to look at a smaller interval sub-J sub-interval sub-J where Z will be defined again linear ODE's have global solutions non-linear ODE's have local solutions as a rule. So now the third step is what we have done so far step 1 we have determined X TS Y TS second step we have determined Z TS now we need to relate Z TS and X TS Y TS we hope that Z TS is U of X TS Y TS so that is what is to be done here in step 3 define a candidate solution at the end of steps 1 and 2 we have a family of characteristic curves which pass through points of gamma okay the family is indexed by the point on gamma for example yes so first of all we solved like this we had a gamma 2 we took a point and passed one characteristic curve through that another point maybe another characteristic another character like that and after solving X TS Y TS now we also have Z TS and what is the initial condition at T equal to 0 it is H S therefore we have the family of characteristic curves are now passing through points of gamma base characteristic curves are passing through the points of gamma 2 and they pass through the points of gamma each member of the family is described parametrically by three typical functions X TS Y TS Z TS defined for T belonging to a sub interval of R thus we seem to have a surface surface it depends what your definition of surface is but we seem to have a surface let us be very vague here described parametrically by two parameters T and S right X TS Y TS Z TS T and S are varying so what you hope to get your surface we know that Z TS was designed to be the value of this that was the definition of Z TS is U of X TS Y TS where U is a solution to the PDE U was not known we thought there was a U and then Z TS we set as U of X TS Y TS then we got an equation for Z thinking that U is a solution to the PDE we did all that so U still has to be retrieved that is what we are saying here define a candidate solution the question now is how to express a surface described by two parameters T and S as the graph of a function U defined on some domain D of course D will be a subset of omega 2 for obvious reasons which is also a solution to the given PDE so how to catch hold of such a function whose graph will be equal to the parametric surface that we have got so a note that a curve or a surface this is very general thing note that a curve or a surface which is parametrically described using smooth functions need not be a regular or smooth curve or surface so it can deceive us it may have a very good description in terms of S and T but actually it is a bad surface it may have singularities we will see that in many examples that we are going to consider in this course the question posed above will be answered not now but in a forthcoming lecture in the more general setting of a quasi linear equation we will do that with full details now we have done for linear and semi linear equations we use base characteristics and then got characteristics and then from there we had a two parameter surface right that what we think is a surface described by two parameters so why not do the same thing here for quasi linear therefore we ask the same questions while trying to solve Cauchy problem for linear and semi linear equations in step 3 we obtained a parametric surface at the end of the step 3 recall that the parametric surface was described using the parameters S and T the parameter S runs through the datum curve S is indexed by the datum curve and for each fixed S as T varies where is the T varying the point will vary on the characteristic curve that is passing through the point f s g s h s therefore the following two questions arise naturally what are they does the parametric surface correspond to an integral surface in other words the question is z equal to u x y where you use a solution to the PDE defined on some domain if the answer is yes okay sometimes you may say that yes it is possible now the question is if the answer is yes then how to determine such function how to determine that solution which defines the integral surface in fact this question can be more general I need not write integral surface does the parametric surface correspond to surface z equal to u x y for some function u then I ask whether that function u is a solution to the PDE whenever we see z equal to u x y and use also PDE such a surface is called integral surface now we have an assertion here integral surface may be constructed using the characteristic curves that is the theorem let D be an open and connected subset of omega 2 let U be a function defined on D which is a C1 function I am not saying is a solution to the PDE or anything it just a solution it is just a function and consider this surface z equal to u x y then the following two statements are equivalent what is that first one the surface is an integral surface of the equation Q L in other words what this statement says is that U is a solution to quasi linear equation Q L second one the surface yes is a union of characteristic curves for Q L so before we proceed to prove the result we need to understand what this previous theorem is about and what it is not about theorem answers the following question given a surface I give you a surface z equal to u x y which is defined by a function which is C1 function defined on some domain D when is it an integral surface of Q L that is the that is what the theorem answers theorem also suggests that an integral surface may be constructed using the family of characteristic curves the small missing point is that that U how do you catch hold of a function U that is where the twist lies let us prove one implies two what is one one is the surface is given to be an integral surface I want to show that is union of characteristic curves therefore what do you mean by this surface yes is a union of characteristic curves for Q L what is the mathematical formulation of this it means that S is a union of gamma P as P varies in S of course gamma P is a characteristic curve passing through P ABC is in C1 of omega 3 their smooth functions even functions therefore a unique characteristic passes through every point in omega 3 since S is a subset of omega 3 we have S is a subset of this union okay take any point in S that guy is going to belong to gamma P for some P in fact the same P let P belongs to S then gamma P passes through P so therefore this is a very simple containment this way of showing so what remains to prove is S contains this union so we want to show this how do we show this take somebody on the LHS how somebody on the LHS looks like he looks like a gamma P for some P in S so let P belongs to S take the gamma P what is gamma P it is a characteristic curve passing through P now we want to show that gamma P is contained in S it is on S what is S S is given by the third coordinate Z is equal to U of the first two coordinates X and Y U of X Y so we have to show that but recall that gamma P is the trajectory of the solution X D by T Z to the characteristic system of ODE is satisfying the initial conditions at T equal to 0 it passes through the point P so X 0, Y 0, Z 0 is P and defined on an interval J at this stage we do not know how big the characteristic curve is except that such a curve exists through P and nearby P so we would like to prove that the entire characteristic lies on the given integral surface S that is the third component is function of the first two components so Z T is equal to U of X T by T this holds for every T in J we would like to show this but this equation is meaningful if and only if this X T by T belongs to the domain of U okay domain of U is already fixed U is given to us it is defined on a domain D therefore it is meaningful only if X T by T belongs to D for every T in J whether it is true or false is a different issue first whether equation makes sense what we question we are asking is it legitimate question or not so U is defined only on D that is the reason therefore entire characteristic curve through P may not lie on S see characteristic curves are coming by solving characteristic equations whereas U is given to you so they do not talk to each other therefore we cannot assert that the entire characteristic curve lies on S because this may not be make sense what we can ask is for whichever T for which X T by T belongs to the domain D then does it hold that is a good question however X 0 Y 0 is in D therefore by continuity X T by T will be in D and you can find a sub interval of J such that X T by T belongs to D fine X T by T itself is defined on J but at 0 you are in D D is an open set therefore for some time T in some interval J dash which is a sub sub sub interval of J U will be in D thus this equation is meaningful for all T in that sub interval J dash note that J dash could be equal to J we are not denying that but what we are saying is that we cannot assert that we cannot assert that J dash equal to J it can be equal if you are not careful in observing this point 1 and 2 we would have proved that entire characteristic curve lies on S this is a place where one makes mistake because we are not careful because what we are going to do later is simply differentiate this equation itself may not make sense if you ignore that you simply differentiate go through the procedure and show that S entire curve lies on S which is not correct it is false in general you have to be very careful okay so we are going to show that a part of the characteristic curve gamma p lies on S that is Z T equal to U X T by T holds for T belongs to J dash fine so define this function this very much makes sense for T in J dash because J dash was chosen such that X T by T is guaranteed to be in D which is the domain of U so this is a good function it makes sense what we want to show is this function is 0 what is the usual strategy show that the derivative is 0 and at some point it is 0 therefore it is identically equal to 0 if a function f dash is 0 and f at some point is 0 f will be 0 because f dash is 0 means f is constant but we are not that much lucky but we can still show that V is identically equal to 0 how do we show this we show that V is going to satisfy a certain initial value problem for ODE okay and it is known meaning that initial value problem which V is a solution to has only one solution it is also known that 0 is a solution therefore V T must be 0 solution that is a strategy we are going to do we will show that V is a solution to initial value problem we will construct or we will design that initial value problem and the initial value problem has a unique solution and 0 is a solution to that and the way we have constructed initial value problem shows that V is already a solution we are also showing that 0 is a solution therefore V must be identically equal to 0 and we have shown that Z T equal to U X T by T holds for T and Z dash that means 1 implies 2 is done so how do you find that initial problem we want this to be solution to that therefore we start differentiating V and get that first order ODE so we will find ODE satisfied by V simply by computing V dash I am not going to this details you have to use just chain rule it is Z T minus U X T by T so Z T derivative is Z dash T this is U X and X dash U Y and Y dash but you know Z dash is C you know that X dash is A and Y dash is B and I want a ODE satisfied by V T so right hand side also I would like to see V T and anything else known functions which are X T by T Z T these are known functions A B C are known functions U is an of course a known function so I write this like this so there is a V here V here and V here so it satisfies the equation U dash equal to F T U I am going to write what is F T U so this is what I want to think as F of T V what is that it is going to be C X T by T U plus this U of X T by T minus this is a known function I will keep it as it is wherever V T is there I put a capital V or capital U because the way I am going to write is in terms of U so this is a function recall that we are interested in applying Cauchy-Lipschitz Picard's theorem and conclude that the V which we have defined to be identically equal to 0 we require the right hand side function F of T U to be Lipschitz with respect to U to apply Cauchy-Lipschitz Picard's theorem so is F the function U going to F of T U is it locally Lipschitz or not that is very important indeed there exists a delta positive such that F is continuous on J prime cross minus delta, delta so the right hand side is continuous function and U going to F of T U is locally Lipschitz continuous in the variable U that is what we need for applying Cauchy-Lipschitz Picard's theorem uniformly with respect to T in Z dash and this follows from the smoothness of the ABC, ABC or C1 of omega 3 and U itself is a C1 function and P is a point in omega 3 because of that we get such a existence of such a delta and I leave it for you to fill in the details. V has the property that V0 is 0 we know that what is V0 Z0 minus U X0 Y0 Z0 is HS X0 Y0 is FSGS so P is in S therefore it is true thus we are led to the initial condition for the above ODE that is U of 0 equal to 0 so this is the initial value problem U dash is FTU where F was described there and U0 is 0. This has a unique solution because F we have shown it is locally Lipschitz therefore this initial value problem will have unique solution. Zero function solves very obvious you have to check that 0 dash derivative of 0 is 0 therefore it is enough to check F of T, 0 is 0 which is verified here F of T0 you have to substitute for U is 0 you get this and that is 0 since U solves quasi-linear equation we are using that because statement 1 is given that U is a solution that is it so this shows 1 implies 2 now we go to 2 implies 1 so we assume that the surface is union of characteristic curves and we want to show that this U is a solution to quasi-linear equation. So let us proceed to prove that U solves QL so take a point X, Y and D and let P denote this point X, Y, U of X, Y that will be on S function U satisfies QL if and only if this equation is satisfied UX, UY minus 1 dot AP, BP, CP is 0 this holds since U since S is a union of characteristic curves and P is in S it follows that there is a characteristic curve gamma P passing through P gamma P is contained in S since normal to the surface S at the point P is in this direction UX, UY minus 1 Z equal to U of XY, UX, UY minus 1 will be a normal direction and ABC at the point XY, UX, Y is direction of tangent to gamma P at P therefore their dot product must be 0 so this shows 2 implies 1. So this is a picture as usual this red X axis green Y axis blue Z axis we saw this earlier UX, Y equal to sign with a solution to the equation UX equal to 0 integral surface is blue it is a union of characteristic curves characteristic curves or straight lines we saw that already they were obtained as intersection of 2 surfaces so datum curve is in magenta so this is the datum curve so that is 0 S sin S as S varies that is the datum curve. Let us summarize what we did we attempted solving Cauchy problems L and QL we were stuck where we had to find that function we stopped there. So, we identified some point that require careful attention there and then we understood the connection between integral surface for a Cauchy linear equation and corresponding characteristic curves. We hope that this understanding will help us in solving Cauchy problem for Cauchy linear equation. Thank you.