 Our journey of studying Cauchy problem for general nonlinear equations has started in the last lecture that is lecture 2.10. Today we continue. The outline for this lecture is first we recall once again the hypothesis and notations involved in general nonlinear equations and in the last lecture we started our search for a characteristic direction we were not at successful. Today the search resumes and we will find a characteristic direction for the general nonlinear equation and then resulting characteristic ODEs are incomplete we are going to see that and therefore we need to extend the system of characteristic ODEs to a new system which is system of equations for characteristic strips. So, let us recall the notations and various hypothesis which are involved in the Cauchy problem for general nonlinear equations. F is a function which is going to define as the general nonlinear equation the arguments of F have been written as x, y, z, p, q. So, defined on for a phi tuples so lying in omega phi which is an open subset of R phi. Such an F defines the most general form of a first order nonlinear general nonlinear PDE by f of x, y, u, u, x, y equal to 0. Though we know that this the this class GE contains in it linear equations, semi-linear equations and quasi-linear equations but we refer to this kind of form of general equation as fully nonlinear equation or general nonlinear equation. So, hypothesis on the function F which defines the PDE or that F must be a C1 function on its domain of definition which is equivalent to saying that the function along with all first order partial derivatives of all the arguments which are x, y, z, p, q are continuous functions on omega phi. And we need to assume that Fp and Fq do not vanish simultaneously at any point in omega phi. Omega 2 and omega 3 denotes the projections of omega phi to x, y plane and x, y, z space respectively. The Cauchy data is prescribed by gamma x equal to Fs, y equal to Gs, z equal to Hs for S belong to an interval I and these functions are C1 functions on the interval I. And we assume the projection of gamma to x, y plane which is denoted by gamma 2 is that gamma 2 is a regular curve which means F primes and G prime do not vanish simultaneously at any point on the gamma 2. Recall that Cauchy problem means we need to find a solution of the equation which satisfies this U of Fs Gs equal to Hs. In other words Fs Gs Hs belong to the surface z equal to Ux, y. Now let us recap from the last lecture the search for a characteristic direction. So, we wanted to extend the ideas from Cauchy linear equations to general equations. The idea in Cauchy linear equations was construct integral surface using characteristic curves. To think of characteristic curves we need to have a characteristic direction. So, how do you obtain a characteristic direction for GE that is a question. Recall for Cauchy linear equations the equation gave us a characteristic direction. The equation QL is AUX plus BUY equal to C. What we observed is if you take any point on integral surface given by z equal to Ux, y. The normal direction at P is Ux, Uy minus 1 and the equation tells that ABC is a direction in the tangent plane at P. This observation led us to define the notions of characteristic system of ODE and characteristic curves. Once again there is no automatic choice of such a characteristic direction for general equations. So, then we went on to consider this family of possible tangent planes to possible integral surfaces through a point for general nonlinear equation. Take a point P naught in omega 3. We get a one parameter family of possible tangent planes given by T1 and T2. What is T1? T1 is simply equation of a plane passing through X naught, Y naught, Z naught. But by restricting FP to satisfy T2 will mean that this is going to be a tangent plane to a possible tangent possible integral surface. It should satisfy these conditions T1, T2. Note this you just the only information coming from the equation. We are not pretending that we know the integral surface that is not needed. Now, we observed that the T2 for quasi linear equation is this and as one of them is nonzero we know that ARB has to be nonzero by assumption in the quasi linear equations. We assume B nonzero then we can solve Q as a function of P. In fact, it is a linear function of P that is what makes things much simpler and T1 for QL becomes this after substituting for this. So, this is a family of possible tangent planes. We have explicitly got one equation. So, this represents the Tp represents one parameter family of possible tangent planes indexed by a parameter. For general equations we have to have both the equations. If you are able to eliminate P or Q from here that is express Q as a function of P go back and substitute here then once again you have only one equation like this. In the case of quasi linear it was very easy to express Q in terms of P. But in case of general non-zero equations it is not clear. So, this is still a one parameter family of possible tangent planes at the point P naught. Now, envelope of one parameter family of tangent planes for QL we computed it turned out to be this. It turned out to be this. What is this? This is nothing but the characteristic direction at P0. This is a line passing through the point x0, y0, z0 and the direction ABC is a characteristic direction. So, therefore the plan for getting a characteristic direction for GE is like this. Obtain an envelope of one parameter family of possible tangent planes for GE which is this. So, get the envelope of this and choose a direction in the tangent plane for the envelope. This plan will be successful because we have a lemma that we proved in the last lecture which is here. If you take a parameter one parameter family of surfaces given by z equal to g xy lambda and you find its envelope which is this z equal to g xy g xy and c lambda denotes the intersection of the envelope and the family then look at the second conclusion. The envelope and s lambda touch each other that means wherever they intersect which is along c lambda the intersection is precisely c lambda at every point on c lambda they share the tangent plane. Therefore, if you can find a direction in the tangent plane for the envelope we are done that will also be a tangential direction for s lambda that will also be yeah s lambda. So, let us resume our search of characteristic direction. So, this is t1 and t2 that represent one parameter family of possible tangent planes at the point p0. Let us find the envelope that is what we need to do. To do that we need to assume that q can be expressed as a function of p in a differentiable way so that this equation is satisfied. In other words we are solving f of x0 y0 z0 pq equal to 0 and q is expressed in terms of p. Question is is it possible? It is always possible to find such a function locally. Implicit function theorem tells you come with a particular solution and nonzero or invertible derivative I guarantee the existence of such a function. This is typically what implicit function theorem says. Now we will go to the implicit function theorem with a particular solution with that means you first find out a p0 q0 real numbers such that f of x0 y0 z0 p0 q0 is 0. That means we have the particular solution. Now we need to see what is this nonzero derivative and fq at this point x0 y0 z0 p0 q0 is nonzero. Please note when we are applying implicit function theorem x0 y0 z0 is fixed. So it is only as a function of p and q that we are trying to solve this equation 1. So fq is nonzero if you assume then implicit function theorem guarantees existence of such a function but locally that means in an interval containing p0 you can express q as a function of p. That is the implicit function theorem. Therefore the one parameter family of planes becomes simply one equation now because I have solved q in terms of p and I put it here. Now only thing to do is differentiate this equation with respect to p and we get this equation these two equations. You differentiate this equation you get 2a, differentiate 2tp equation you get 2b. From 2a and 2b you can eliminate q prime p and we get this x minus x0 y minus y0 is proportional to fp fq okay something we got. Now what is this p0 pqp p0 actually x0 y0 z0 f is a function of 5 variables right we need a 5 tuple here for want of space here I have just made it a small notation p0 it stands for this. Using this and the equation for tangent planes which is tp here we get the curve of intersection and that is this okay x minus x0 is proportional to fp y minus y0 is proportional to fq therefore z minus z0 is proportional to pfp plus qfq that is what we have written here. So this is a line right cp is a family of lines of course indexed by p the same thing was actually one single line it never depended on p for a quasi linear equation all of them passed to the point x0 y0 z0 and having a direction fp fq p of p plus qfq the 3 tuple which is given in 5 is non-zero because we have assumed that fp square plus fq square is not equal to 0 throughout the domain of f that is a hypothesis on it. An envelope of the family of planes tp by definition is a union of cp that means union of these lines so envelope is a cone having vertices at p0 so visualize the way a family of lines passing through a point p0 for example you have this point p0 right so imagine lines in 3d okay like that so that is how looks like a cone this is called monge cone thus we have a monge cone field defined in omega 3 so omega 3 is in R 3 some set okay at any point you have a cone sitting there at another point you have another cone like that so this so every point you can attach a cone. So this is called cone field similar in spirit to a vector field to any point if you attach one vector associate one vector it is called vector field if you associate cone it is called cone field that is to every point of omega 3 a monge cone is associated for quasi linear equations the monge cone degenerates into a straight line because we observed that the envelope is actually a straight line it is independent of p so therefore just one single straight line so the monge cone field reduces to a characteristic vector field in the case of quasi linear equations. Now each of the lines Cp corresponding to each fixed p is a generator of the cone generator of the cone means it is a line in the cone so you have a for example you have a cone like that you have a line here so at a point on the cone if you take tangent plane okay that line is going to be there on the tangent plane therefore what we wanted was to look at the envelope and take one direction in the tangent plane of the envelope now here it is very nice tangent plane of this cone at these points the one of the direction is clearly the line Cp so we can take that Cp to be characteristic direction so a characteristic direction for GE is found for every p in omega 3 and every pq we do not we no longer require q of p here pq after all q is a function of p q is some value right it satisfies this equation still we have found a characteristic direction at p what is that fp fq p of p plus q of q now we have to look at the characteristic odys we are going to say they are incomplete we will see why so inspired by the success in finding an integral surface using characteristic curves we hope to repeat the same for GE also in ql we were successful so we hope same thing happens for GE as well so we are in search of curves in omega 3 which have tangential direction equal to the characteristic direction that we just found which is this fp fq p of p plus q of q at each of its points p p such curves we are looking for now pq of course depend on the point x y z right they satisfy this equation so it depends on x y z let us write a curve having a characteristic direction so let j be an interval 0 belongs to j gamma p0 be a curve given by xt by dz t as t varies in j such that it passes through the point p0 at t equal to 0 and it has a tangential direction which is characteristic direction at each of its points this is what we would like to call a characteristic curve therefore the following system of characteristic odys hold along gamma p0 so dx by dt dy by dt dz by dt are fp fq and p of p plus q of q and at t equal to 0 we want to be at the point x0 y0 z0 so fine now why is it incomplete I do not know what is p and q that is a problem okay let us denote the above system by Kara word e exactly like we used for quasi linear equations the solutions of this its images images will be characteristic curves only thing is that this is not it does not determine characteristic curves that is the problem because there are pq's here it is not solvable as unknowns pq appearing in it depend on xt by dz t via this equation f of xt by dz t pq equal to 0 p and q are also depending on t thus we also need to determine p and q as functions of t since xt by dz t themselves are unknowns there is no hope of solving for p and q from this equation and note this is not the case for quasi linear equations because characteristic odys never involved p and q and this is a second difficulty in extending the ideas from the quasi linear case what is the first one finding a characteristic direction that was the first difficulty we have overcome and second difficulty we will overcome by supplementing two more equations for 1 for p and 1 for q so how to find equations for pt and qt we need to supplement Kara word e system with equations for pq along the curve xt by dz t and what should be the property pt qt minus 1 should be the normal direction to a possible integral surface they are not arbitrary functions so we have to determine five quantities xt by dz t and pt qt and not just the three quantities because we are unable to determine the three we would have been very happy if we would have got xt by dz t we could have proceeded but the equation for xt by dz t involved p and q therefore we need to find p and q also so geometrically speaking what we are saying is we have to determine not only a curve but also a tangent plane to a possible integral surface that contains this curve so in other words we are trying to find a curve like that so suppose this is the point xt by dz t so we want to find something like that pt qt of course minus 1 I write but then it is okay minus 1 there is nothing to find so we need to find this so let J be an interval in R and gamma be a curve given by x equal to xt y equal to y dz equal to dz t be a curve having tangential direction as char dir that is characteristic direction at each of its points where pt qt satisfy this equation f of xt by dz t pt qt equal to 0 the phi tuple xt by dz t pt qt is called a characteristic strip maybe you may put like that because we are saying tuple but even without that it is fine all these five together is called phi tuple that is called a characteristic strip and the point xt by dz t is called support of the characteristic strip because there is a point at which you are putting a plane with normal pt qt minus 1 so let us illustrate that pictorially so here there are two points I have considered index by t equal to t0 here and t equal to t1 so x t0 y t0 z t0 is a point this is a tangent this is a plane with the normal pq minus 1 this is another point where the normal is pq minus 1 so it can be strip can be thought of as a point along with an infinitesimal plane element passing through that point xt by dz t with normal direction pt qt minus 1 now how do you get equations for pt qt so here we assume some more nice properties about the solution u see u is supposed to be a solution to first order pde we have not yet found but we are assuming that it is c2 is it okay that is a question it is okay because we are going to use this only to derive certain equations and for deriving equation you assume what you want does not matter but after getting the equations then you should show that solution exists there you should not suppose that c2 etc okay we will comment on this in the next lecture also so this is only to derive the equations for pt qt that we are assuming u is c2 so how do we derive the equation for pt what are at our hands this equation f of x y u u x u y equal to 0 differentiate that with respect to x so first is fx here I am using the zeta to stand for the x y z pq okay so fx and then with respect to z in z you have u x y therefore you need to differentiate fz and u with respect to x then with here also p fp and then whichever is here with respect to x which is u x x and here it is fq and u x y or u y x equal to 0 at every point in omega 5 where we use this notation zeta equal to x y z pq no actually this is zeta is not x y z pq it should be x y we have to substitute x y z pq equal to x y u x y u x x y u y x y that is what it is because we are going to differentiate here this u is a function of x y u x is a function of x y u y is also a function of x y so that is why we get this by chain rule and x t y t z t is a point lying on the integral surface z equal to x y we require the following thing to hold namely p of t I want it to be like u x so p of t is equal to u x of x t y t we are demanding this okay because pt qt minus 1 should be such that f of x t y t z t pt qt should be equal to 0 right so p is supposed to play the role of u x along the curve x t y t so that is my pt we want this so on differentiating the above equation with respect to t and using chain rule we get p prime t is equal to t appears in both the variables so differentiate this with respect to x u x x at the point x t y t into derivative of x with respect to t that is x prime t and differentiate this with respect to y that is u y x at the point x t y t into derivative of y t which is y prime t since characteristic od hold for x t y t z t denoting zeta t equal to this phi tuple x t y t z t pt qt we get p prime t is equal to u x x x prime is f p y prime is f q so we have got this equation for p prime and from the equation 7 that is the one we got after differentiating f of x y yeah this one this equation and here we almost got an equation for p prime the only problem is there is u x x and u y x we do not want that right equation of p should involve only f it can involve activities that you know problem but not u x x and u x so that needs to be removed now so we solve for that which we do not want in terms of what we know f x p is what we want to find f z which we know therefore p prime t equal to this and now becomes p prime t equal to minus f x plus p f z this is the equation for p prime similarly we do for q what do we do we look at the same equation ge differentiate that with respect to y we get something then we propose value for qt qt is supposed to be u y of x t y t differentiate this with respect to t it will involve this u x y and u y y eliminate this using the previous equation I think it equation 11 and then you get an equation for q prime x is replaced by y that is the only change p is replaced by q so this is the equation for q prime so we got the equation for q prime also so this is the system where the first three we have obtained or we have proposed using characteristic direction we realize that p q also depend on x t y t z t and we said we have to find the equation for p t qt and we have got we have appended Kara ODE with these two equations now it is called Kara strip ODE equations for the characteristic strip now there should be no problem because x y z p q here also x y z p q anything else f we know f therefore we know f p f q so no problem of course this is a non-linear system of ordinary differential equations how to solve this we have to solve this to get the characteristic strip and that we will do in the next class so let us summarize what happened so far we saw that the quasi linear equations has a characteristic direction right away given by the equation q l g e gives only possible tangent planes and using the idea of envelopes we found characteristic directions for g e and we observed that the system Kara ODE is incomplete so the system is supplemented with two more equations and we got characteristic strip equations ordinary differential equations for characteristic strip so in the next lecture we are going to take up the Kara strip ODE system which is a system of 5 equations non-linear equations and we need to solve them and what is the strategy in quasi linear equations we take the datum curve take a point on the datum curve and pass a characteristic curve through that so now if you take datum curve what is known is only x y z values on the datum curve namely f s g s h s which are given to us we will use them as initial conditions in the characteristic system of ODE in the quasi linear case but in the general non-linear equations case we have two more equations which is p t and q t equations for them therefore we should know on the datum curve what should be the values of this p t q t on the datum curve so that is what is called initial strip so initial data or datum curve is given that must be extended to a strip and that is called initial strip using initial strip and using the initial conditions coming from the initial strip we will solve characteristic strip ODE's and get characteristic strip we will determine characteristic strip and from there will come characteristic curves and from there we try to take their union and get the integral surface so these steps will be implemented in the forthcoming lectures thank you