 In this lecture, we are going to complete the proof of local existence and uniqueness theorem for Cauchy problems for the general nonlinear equations. So we start with a recap of what happened so far and then we go to step 3 with namely defining a candidate solution. So we will define a candidate solution by applying inverse function theorem and then we show that the candidate solution is indeed solution. So this is to recall the notations QL stands for quasi linear equation and GE stands for general nonlinear equations or sometimes people call it a fully nonlinear equations. Of course, GE contains L, SL as well as QL but when it is presented in this form it is called generally nonlinear equations. Where are we in solving the Cauchy problem? The key steps involving the solution of Cauchy problem were identified. Step 1 is to obtain a system of ODEs for the characteristic strip and step 2 is a finding initial strip. Third one is to define a candidate solution and fourth one is establishing that the candidate solution is indeed a solution. So steps 1 and 2 were successfully implemented so far and let us recall the difficulties once again which are new to GE when compared to QL and what were the ideas that helped us to overcome them. So our analysis was motivated by quasi linear equations QL. So QL gave us characteristic direction, characteristic direction gave rise to characteristic curves, characteristic curves made up an integral surface. Now equation GE does not give away a characteristic direction. So a useful idea we observed that for quasi linear equations the characteristic direction is the envelope of possible tangent planes. In fact, the possible tangent planes envelope is a straight line which has the characteristic direction. So we found that the same idea works for GE as well. So characteristic direction at a point PXYZ is given by Fp, Fq, PFP plus QFq. The argument has to be 5 to pull P, Pq capital P stands for XYZ, small p and small q they are such that capital F of this is 0. And characteristic ODE system is incomplete for general nonlinear equation because when we try to find a curve which has a characteristic direction it has to satisfy this set of ODE's namely Kara ODE but here p and q appear which themselves are unknown. So this is not solvable and we need to add equations or supplement the system with another two equations one for p and one for q. It has been achieved and that was called Kara strip ODE where the dp by dt dq by dt equations have been added. Now next question is that can we pass characteristic curves through points of the datum curve this is what we did in QL. So for that we need to solve Kara strip ODE with initial conditions given by points of the datum curve initial conditions for XYZ are given by the datum curve no initial conditions are known for p and q. So they were found in step 2 which was called finding an initial strip. Next step is to solve the initial value problem Kara strip ODE plus an initial strip that will give you a candidate solution it will define a candidate solution we are going to see that. So from now onwards what we assume maybe in deriving these equations we have assumed certain higher smoothness conditions on FGH and the capital F or maybe U but do not worry. Now onwards we assume this that characteristic strip ODE is given equations are given to you you do not have to derive and initial strip consisting of C1 functions is found. Of course we cannot find perhaps throughout gamma but you look at a point FS0 HGS0 HS0 on gamma and hence I want this to be defined nearby that point p0 and let this functions XYZ pq which are functions of t and s solve the IVP. What is IVP Kara strip ODE supplemented with this initial conditions coming from initial strip. Now defining a candidate solution this is where we need to apply inverse function theorem to certain functions and then get a candidate solution. So we are interested in the invertibility of this function. What is this function this is actually the base characteristic curves right the trace is base characteristic curve as t varies passing through the point FSGS which is on gamma 2. So near this point 0S0 which belongs to J cross i. So now to apply inverse function theorem the Jacobian is required to be non-zero. Of course we need to check whether phi is C1 that is C1 because X and Y are solving ODE's therefore the derivative with respect to t is continuous no problem with respect to x it is also differentiable and C1 because of the differentiable dependence. So we have the C1S of phi now we need that the Jacobian is non-zero the Jacobian is this as before we do not once again analyze with g of ts because nothing is known for a non-zero t. So only at 0 it is known j of 0S0 the Jacobian is this which is f p f q at the point zeta 0 and f dash S0 g dash S0. Of course the way we got pS and qS is such that p of S0 is p0 q of S0 is q0. So you could as well write here pS0 qS0 but remember we started with p0 q0 a particular solution of the system of equations which define pS and qS later on. So you can write pS0 qS0 or p0 q0 because both are actually the same. Now the Jacobian condition is precisely the transversality condition f prime g prime corresponds to what tangent to gamma 2, tangent to the gamma 2 what is gamma 2 projection of gamma 2 x y coordinate x omega 2 this is the tangent this is the base characteristics tangent to the base characteristics of this direction f p and f q we have to assume is non-zero that means they are not parallel. So base characteristics cut gamma 2 that is the transversality condition this is same as the delta naught that we saw in the step 2. So defining a candidate solution by inverse function theorem there exists an open subset E of j cross i containing the point 0, S0 and f which is an open set of omega 2 which contains x of 0 S0, y of 0 S0 which is actually fS0 gS0 and a continuously differentiable function from f to E such that these two compositions are identity maps on respective spaces. This is a picture which depicts so this is the phi map we went inverse function theorem told that there are neighborhoods here E and f and a map psi is defined this where we are using TS coordinates here we are using x y notation phi restricted to E is a diffeomorphism and so on and denote psi of x y equal to t of x y S of x y and the equation psi circle phi is identity on E phi circle psi is identity on f they give t equal to t x y S equal to S x y. So recall that jet TS was expected to be the value of the solution at the point x TS y TS therefore this motivates us to define a candidate solution by using this z. So u defined on f to r u of x y equal to z of t x y S of x y as a function u is a composition of two C1 functions by chain rule u is selfie C1 function on f. So step 3 is successfully completed we have defined a C1 function as a candidate solution in the next step we are going to check that this is indeed a solution. So this is a solution to the Cauchy problem if the following identity holds for every x y in f that is capital F of x y u x y u x x y u y x y equal to 0 and the Cauchy condition u of f s g s equal to h s is satisfied. So define i dash to be those values of s in i for which f s g s belongs to f. So this is that i dash which came there. So s belongs to i dash you will be the datum curve will be on the corresponding integral surface defined by this u. So observe that t of f s g s is 0 and s of f s g s is s. Thus for s in i dash we have u of f s g s equal to by definition of f s u of f s g s u of any two quantities is t of f s g s comma s of f s g s. But that is nothing but z of 0 comma s which by the definition of z is h of s. Therefore Cauchy condition is satisfied that means a piece of the datum curve lies on the integral surface corresponding defined by this function u. Now we have to still check that u solves the PDE. So proving that for every x y u satisfies the PDE namely f of x y u x y u x x y u y x y equal to 0 holds is the same as showing that for every t s in E f of x t s y t s z t s u x of x t s y t s u y of x t s y t s is 0. Because x y and t s are in one to one correspondence via difiumorphism. Therefore showing this is same as showing this. And we already know that for t s in J cross i f of x t s y t s z t s p t s q t s is 0. This we have seen in step 2 while defining initial strip we read that. So this is now we want to show this. There is a difference between the two. The difference being in the last two coordinates u x x t s y t s is here, p t s is here, u y x t s y t s is here, q t s is here. Suppose we show that this pair of functions of t s is same as this pair then we have shown this because it is already known. So let us try to do that. Let us show that p t s is u x x t s y t s, q t s is u y x t s y t s. So this is the triple star is what we want to show. Then this holds. This we already observed on the last slide that completes the proof. So what remains to show is these two equalities. How are we going to show that? We prove that this pair is this pair or same by showing what? We will show that this pair is a solution to a system of non-homogeneous linear equations. This is also a solution of non-homogeneous linear equation. The same system. System has a unique solution because the coefficient matrix there will be invertible. Therefore the solution must be the same. We know that the system a x equal to b, if you have a x 1 equal to b and you also know you are at a x 2 is equal to b that would imply x 1 equal to x 2. If you know that a is invertible. So solution is unique. So we are going to show this. We are going to derive this system of linear equations which both the pairs satisfy and we will show that the coefficient matrix is invertible. Therefore the solution is unique. Therefore if you knew a x 1 equal to b and a x 2 equal to b then it must be that x 1 should be equal to x 2. And then it will follow that this pair, this pair is same as this pair which is triple star. So now we have to get those system. How do we get that system? So differentiating this equation z t s equal to u of x t s y t s with respect to s and t will give us two equations for convenience I write in the matrix form. What is z s? z s that is the first equation. z s means u x into x s. That is u x into x s plus u y into y s. So y s into u y. So that is the first equation. This one, this into this. This is a matrix, this is a vector, this is a vector. So this equation is very clear. Now we claim, so what is this equation? This is satisfied by u x x t s y t s, u y x t s y t s. This matrix acting on that will be z s t s z t t s. Now we are going to show p t s q t s also satisfy same system. See the coefficient matrix is same. In this case I have written as a left hand side. So this is same. This is same. And if this is invertible these two must be same. Is this invertible? That is a question. It is invertible because change of variables. Change of variables, this is a Jacobian corresponding to change of variables. Therefore this will be always invertible. So therefore the moment we establish this system it automatically follows that this pair is same as this pair. Now how do I derive, this 5 actually stands for this equation. How do I derive this is missing in the latex compilation it has vanished. The second equation follows from Karash Tripodi because that is what it is. Z t equal to x t p plus y t q. What is x t and y t? They are f p and f q. So it is p of p plus q of q. Therefore this follows. So we have to show the first equation. The first equation is this we want to show. So we want to show that this thing equal to 0. Let us call it by ATS. We want to show that ATS is 0 for every TS in E. How are we going to show this? Because this features derivatives with respect to S that may be the difficulty. For each TS we show as a function of t it solves an initial value problem A t plus f z A equal to 0, A of 0 is equal to 0. We will show that it solves this initial value problem. Now by uniqueness of solutions to initial value problem ATS must be 0 for all t. Why? Because this initial value problem will have only 0. Zero solution is a solution? A of t equal to 0 is a solution? Here it satisfies this condition? Therefore if this is a linear equation. If this is continuous coefficients this will be lift shifts. Therefore you have a unique solution. Therefore ATS will be 0 for every t and this happens for every fixed S. Therefore ATS is 0 for all TS. Fine. So we have to derive this equation that is all remains to show. A of 0 S is 0 for all S. This is coming from the definition of initial strip. H prime equal to P of prime plus Q of Q g prime. Differentiating ATS we have to derive an equation satisfied by ATS. So the only thing you can do is differentiation. So A t equal to z s t at P t x s Q t y s P x s t Q y s t. Now a small rearrangement is required it is an algebra so that we will get f z into A minus f z into A. So we are going to use Kara strip ODE equations here and we end up getting this that is same as minus f z A because this is 0. So we use the dou by dou S of f is 0 which is a consequence of f of x y z P Q being 0. So note that the coefficient matrix appearing in the system of equations 4 and 5 is the same we already observed. Its determinant is precisely this. It is the Jacobian corresponding to the change of coordinates x y and t s therefore it is always non-zero. Thus the linear system 4 5 has a unique solution. Thus we have proved the following local existence and uniqueness theorem. Uniqueness proof proceeds as in the case of quasi-linear case so I am not going to do it here. So what is the theorem? Assumptions on f omega 5 is an open set connectedness is not required open set. Let f be a C2 function that we cannot dispense with. F p and F q satisfy that both of them cannot vanish simultaneously at any point of omega 5. Assumptions on Cauchy data, Fgh are C1 functions so no need of C2 functions just C1. And assumption on initial strip and transversality condition. So assume that this system f s g s h s p s q s equal to 0 p s f dash plus q g dash equal to h dash admits a solution p s q s where p s q s are continuously differential functions on the interval i. Actually we have it is enough to work with the existence of these kind of functions for s in a small interval containing some point s 0. Because finally the conclusion of inverse function theorem or implicit function theorem are local. So assume the transversality condition. Conclusions are the general nonlinear PDE admits a solution defined an open subset of omega 2 for the Cauchy problem that is missing here Cauchy problem. The Cauchy problem for general nonlinear PDE admits a solution defined an open set of omega 2 which is f actually it is not d it is actually f we have found d is equal to f the proof in the proof we have d is equal to f and the point f s 0 g s 0 is in f satisfies u f s g s equal to h s for those s for which this is in f further the solution is locally unique. So let us solve an example of where we are going to solve a Cauchy problem for a nonlinear equation the simplest nonlinear equation u x square plus u y square equal to 1 Cauchy data is given by u is 0 on the circle x square plus y square equal to 1. So first thing as always is to parameterize the Cauchy data x equal to cos y equal to sin as z equal to 0 s in the interval 0 to pi system of ODEs. So what is f of x y z p q p square plus q square minus 1. So in this example so we should always write this what is this function from here we can compute very easily f x f y f z are 0 f p is 2 p f q is 2 q. So dx by dt is f p therefore 2 p dy by dt is f q therefore 2 q dz by dt is p of p plus q of q what is p of p plus q of q let us write down once maybe f p is 2 p f q is 2 q p f p plus q f q is equal to 2 p square plus 2 q square but the equation says p square plus q square equal to 1 therefore this is 2 and d p by dt is 0 d q by dt is 0 because our f does not involve x y z at all. So we have to now complete into an initial strip the datum curve we have to find p s q s satisfying 2 conditions one is the equation p s square plus q s square equal to 1 second is p of prime plus q g prime equal to h prime that will give us this. Now we have to find solutions here we see clearly 2 solutions of course you may say infinitely many solutions because you can mix both of them because this is an algebraic or whatever transcendental equation right at some yes you can be here at some other as you can be here you can keep on oscillating but that is not good what we need is a smooth function this is very important because what we are trying to solve just recall whether quasi linear or fully non-linear we take gamma we take a point on this and we pass a characteristic curve through that and repeat for everything right whenever you choose a point on gamma it will correspond to some yes and then you are simply passing a characteristic curve through that another s dash you take you pass through pass another characteristic why should these characteristic curves together stitch or weave a surface that would happen if things underlying are smooth functions that is why f g h we assume c 1 functions. So, here we need to assume p s q s also smooth functions otherwise we do not expect so that is why there are 2 choices for smooth functions. So, p 1 s q 1 s is cos s sin s other one is minus cos s minus sin s so smooth is very important. So, there are only 2 if you insist on smoothness now if you take the initial strip where p q is taken to be cos s and sin s the solution of characteristic o d e will be this I am not going to the computation because it is a very simple o d e is that you can solve. So, these are the solutions x t s y t s z t s p t s q t s now from the first 3 equations we get this relation x square plus y square equal to z plus 1 whole square and z t s is given by this formula on simplification you get that. So, we define a pair of functions u 1 u 1 and u 2 1 with plus sign 1 with minus sign then the solution to Cauchy problem is given by this and not the minus 1 why it does not satisfy the Cauchy data you can check that. So, if you proceeded by taking the initial initial strip as minus cos s and minus sin s then we would have got this as a solution and not this will not satisfy the Cauchy data this will satisfy. So, summary is for this lecture is that the candidate solution was defined we verified that is indeed a solution to the Cauchy problem a closer look at the proof of existence and uniqueness theorem reveals proofs of the existence and uniqueness theorem for both q l and g e are strikingly similar actually the same but for obvious modifications once characteristics tips have been obtained because we need that x t s y t s z t s right and we always work with x t s y t s to get inversion right inverse function theorem is applied only for x t s y t s whether it is q l or g e and the extension of ideas were clearly brought out in our presentation. So, this completes the analysis of Cauchy problem for general non-linear equations in forthcoming lectures we take up some problems based on this. Thank you.