 Welcome to tutorial on Quasi-linear equations. In this we are going to solve some Cauchy problems for Quasi-linear equations and we also explained through examples the local nature of solutions to Cauchy problem. Recall that the existence and uniqueness theorem gave us existence of a solution only nearby any fixed point on the datum curve. So, is that all that can be expected or can we get a solution whose integral surface consists of entire datum curve or defined on entire domain omega 2. These are the questions we are going to discuss under this heading on the local nature of solutions to Cauchy problem. So, let us start some problems examples. This is the simplest example that we considered in the beginning u x equal to 0 and data is u 0 y equal to sin y. So, how do we solve this? We need to first parameterize the given Cauchy data x equal to 0, y equal to s, z equal to sin s and y in r therefore s in r. So, this is our datum curve. Then we need to look at the characteristic system of ODE for the given equation. Recall dx by dt equal to a in this example a u x that is a is 1, b and c are 0. So, dy by dt equal to b which is 0, dz by dt equal to c which is 0. So, this is the characteristic system of ODE associated to this equation. Now, we need to solve these ODE's system of ODE's. So, with the initial condition so that at t equal to 0 we are at a point on gamma. So, x of 0 when time is 0 equal to 0, y of 0 is s, z of 0 equal to sin s because 0, s, sin x is a arbitrary point on gamma that datum curve. Solution is very simple to obtain. So, we get x equal to capital X of t s equal to t, y equal to capital Y of t s equal to s and z equal to capital Z of t s equal to sin s. That is very easy to see because see here dx by dt equal to 1. Therefore, x has to be t plus constant at t equal to 0 x must be 0. Therefore, this is the x equal to t. Here dy by dt is 0 that means y is constant at t equal to 0 it must be s therefore y equal to s, dz by dt equal to 0 that means z is constant with respect to t. But at t equal to 0 it should be sin s that will give us these solutions. Now, we need to eliminate or we need to solve for t and s in terms of x and y from the first two equations which is obvious in this example t is capital T of x, y equal to x s equal to capital X of y capital S of x, y equal to y. So, we have got this. Now, we need to substitute in this and we get the solution. So, u x, y equal to sin y which is defined for all x, y and r2. These are the pictures that we have already seen the red is x axis green is y axis and this is z axis and this one which is in magenta color is the initial data u0, y equal to sin y. So, 0, s, sin s as s varies in r you get this curve and integral surface is a blue color one. These are the characteristic curves here they are straight lines we already saw that. Let us look at another example this is also linear equation u x plus u y plus u equal to 1 and the Cauchy data is of this nature it is prescribed on some curve u of x comma x plus x square equal to sin x. So, as before you need to parameterize the Cauchy data. So, s x equal to s y equal to s plus s square and z equal to sin s and for s positive done and we need to write characteristic system of ODE dx by dt equal to a in this example a is 1 dy by dt is b b is 1 dz by dt is c remember c is anybody who is on the other side. So, it will be 1 minus z you have to be careful there do not think it is 1 it is 1 minus z because equation is of the form a u x plus b u y equal to c. So, dz by dt equal to c therefore it is 1 minus z. Now, we need to solve the system of characteristic ODE with the initial conditions which are given here and we get solutions as this x is t plus s y is t plus s plus s square z equal to 1 minus e power minus t plus e power minus t into sin s fine. Now, we need to find the t and s as functions of x and y using the first two equations x equal to t plus s and y is equal to t plus s plus s square it is not clear to me how to get it but let us ask whether it is possible to get at all. So, therefore, here it looks like it is possible to get. So, t plus s is actually x therefore x s square is equal to y minus x therefore s equal to root y minus x because s is positive that is why I am not taking minus so root y minus x. So, once you know s t can be obtained from here x minus s that is why x minus root y minus x yeah so it is possible to get and then substitute for t and s in this formula for z you will get a solution it is a solution and now question is we have got the formula ask what is the domain on which it is defined first of all you need that y minus x should be positive because there is square root that is it everything else is fine. So, y minus x is positive that is the restriction which is y is bigger than x so that means it defines a solution in this domain x y in R 2 such that y is bigger than x. So, this is the integral surface it is in blue the atom curve is in green. So, along this line y equal to x the formula has a problem right. So, you will see that corresponding trouble here when y equal to x on the line y equal to x. Let us look at another example in this example what is happening is u x plus 3 y power 2 by 3 u y equal to 2 now a is 1 b is 3 y power 2 by 3 this is not a c 1 function our theorem required c 1 function right it is not c 1 function. Let us see what happens c of course is 2 constant no problem Cauchy data is u of x 1 equal to 1 plus x. So, first thing as always is to write gamma in the parametric form we use this and characteristics system of ODE is this now we need to solve this system of ODE with initial conditions this solution is this x equal to t plus s y equal to t plus 1 whole cube z equal to 2t plus s plus 1. Now, from the first two equations we get t equal to y power 1 by 3 minus 1 and s equal to x plus 1 minus y power 1 by 3. In fact, for t you use this equation because it does not have s. So, from here we get t once you get t substitute here you get for s that is why we got this. Now, go back and substitute in this formula x plus y power 1 by 3 and where is it defined? It is defined everywhere but then actually everywhere in R2 because this is just cube root of y right cube root of any variable makes sense but the problem here is that it is not differentiable at y equal to 0. So, we have to choose either y positive or y negative we have chosen y positive and these are the views of the integral surface with the different orientations remember always the axis red is x green is y and the blue is z axis. So, you see that some steeping is happening here around y equal to 0 should happen right because y power 1 by 3 is there it is not differentiable something it should be reflected in the picture okay. Let us look at another example this is a very non standard example this was constructed with something else in mind and but it turned out to be it is a very good example it is an quasi linear equation. So, far we have considered only linear equation this is a quasi linear equation sin u u x plus u y equal to 0 here a and b a is sin u if a is 0 b should be non-zero but b is always 1. So, it is a square plus b square is non-zero fine Cauchy data given is u of 0 y equal to y for all y in R let us all parameterize the Cauchy data 0 s s s in R characteristic system of ODE is this now because of the quasi linear nature equation for x actually involves z now whereas y and z does not involve any other variables. So, from here you can see that along the characteristic curve z is constant because d z by d is 0 and what is d by d t equal to 1 means y equal to t plus constant therefore because at t equal to 0 it should be s y is t plus s z is constant and that constant has to be s therefore you put that s here and integrate this. So, we get sin s into t and at t equal to 0 it should be 0 so this is a solution y is t plus s z is s now using x and y equations we have to get an expression for t and s but you see I do not think it is possible because t sin s is there this is t plus s so this is okay nice t plus s but there is a sin here. So, we cannot express I cannot express explicitly then I ask is it possible for anybody to express at all which means is the inverse function theorem applicable we will check that this was not the case so far in all the earlier cases we could solve maybe it is a bad function of x and y it does not matter but we could explicitly solve here explicitly we are not able to solve fine. So, to know if a solution exists we have to rely on the existence uniqueness theorem now we have no choice. So, in this example we are not going to get explicit form of the solution because we are unable to invert we are unable to write t and s as functions of x y. So, when you look at the whether it is possible at all the j 0 s turns out to be sin s of course sin s is 0 when s whenever s is a multiple of pi okay all integral multiples of pi other than that it is always non-zero. So, if s is equal to k pi for some k integer this Jacobian is 0 if it is not like k pi for some k then it is always non-zero these are all isolated points okay. So, Jacobian if you remember we have pointed out the ways of failure of of transversality condition and there we said it is a possibility that you have a sequence s n along which j is 0 okay but here and converging to some point here it is not happening these are all isolated points okay there is no convergent subsequence of these multiples of pi okay otherwise Jacobian is non-zero therefore local existence and uniqueness theorem is applicable whenever s is not a multiple of pi. So, in terms of y not it is y not is not an integral multiple of pi and we conclude that there exists an integral surface for the given PDE containing p not and a piece of gamma of course question remains what happens when s is an integral multiple of pi that is to be explored. So, now let us look at some examples which illustrate the local nature of solutions to the Cauchy problem. Before that let us let us revise the notion of local solution in initial value problems for ODEs this is the equation we consider d by dx equal to f x y and y x not equal to y not this is initial condition. So, both equation and initial condition together is called initial value problem called IVP in short of course we need to assume something on f let us assume that f is a continuous function. Now a solution to the IVP which is defined on the interval i is called global solution what is i? i is here this ODE makes sense for x in i and if you have a solution which is defined for every x in i we call it global solution. Imagine it is not the case and solution is defined only on a sub interval of i the proper sub interval of i then that is called local solution. Now recall that Pino's theorem and Cauchy-Lipsch's Picard theorem whenever it is applicable always guarantee the existence of a local solution to IVP. They do not talk about global existence there are other theorems about global existence and there is a full understanding of what happens if a local solution can be extended to make it a global solution if you fail somewhere what are the precise reasons why you cannot extend it to a global solution. So, that is very much understood for initial value problems for ODEs but that is not the case in my opinion for partial differential equations I have not come across such results. Let us now define for partial differential equations a solution to a Cauchy problem for a Cauchy linear equation it can be for any equation first order PDE because we are in this first order PDE setup and this is a tutorial on Cauchy linear equations we can as well assume for Cauchy linear equations otherwise concepts are quite general. So, a solution to a Cauchy problem is said to be a global solution if this is where something comes with respect to datum curve if the corresponding integral surface contains the entire datum curve okay you have a solution right then look at the integral surface Z equal to XY entire gamma if it is on that we say it is a global solution with respect to datum curve otherwise the solution is called local solution with respect to datum curve. We have another related notion a solution to a Cauchy problem is said to be a global solution with respect to domain with respect to domain if the solution is defined on the domain omega 2 what is omega 2? Omega 3 is a set on which the Cauchy linear equation was defined right the coefficients A, B, C they are defined on omega 3 projection of omega 3 to XY plane is omega 2. So, you would expect that solution should be defined throughout omega 2 if it is so we are happy and we will call it global solution with respect to domain otherwise solution is called local solution with respect to domain. Recall the existence and uniqueness theorem proved in lectures 2.6 and 2.7 they guarantee the theorem guarantees the existence of a local solution with respect to datum curve and with respect to domain both. Now since gamma 2 is a subset of omega 2 if a solution to the Cauchy problem is global with respect to domain that means it is defined throughout omega 2 it is also defined throughout gamma 2 gamma 2 is a projection of gamma so it should be global with respect to datum curve. Observe that if a solution to the Cauchy problem is local with respect to datum curve then it is also local with respect to domain. Okay now this is a remark applicable for semi-linear equations. Recall that the base characteristic curves are defined as projection to omega 2 of the characteristic curves in omega 3 in the context of general quasi-linear equations. But in a semi-linear equation what happens is that base characteristic curves can be found out independent of the characteristic curves because the equations governing the base characteristic curves namely dx by dt equal to a and dy by dt equal to b involve only x and y a and b are functions of x y only it does not depend on z therefore base characteristics can be found independent of characteristic curves. Now we observed in step 2 namely in the proof of the existence uniqueness theorem there we observed that the equation for z may not admit solutions for all t for which base characteristics are defined simply because the dz by dt is a non-linear equation for a general semi-linear equation dz by dt is a non-linear equation and solutions to non-linear equations as we said as a rule are only local solutions so it can even cut out some portion of this base characteristic curves. So this might result in a situation where projection of a characteristic curve may not be the entire base characteristic curve that you already found otherwise let us assume all these curves are the longest possible things that we have found. We will see an example it will be obvious so the next example exhibits this possibility ux plus uy equal to u square and the Cauchy data is ux0 equal to x. I think this is the simplest complicated semi-linear equation because u square is the first equation that we learn even in ODE dy by dx equal to y square that is the first non-linear equation that we will come across in first order ODE's. So let us parameterize the given Cauchy data x equal to s, y equal to 0, z equal to s, s in R. The characteristic system of ODE is dx by dt equal to 1, dy by dt equal to 1, dz by dt equal to z square. Now when we compute the base characteristics x equal to x of ts equal to t plus s because it is t plus constant it has to be t plus s, y is just t of course z also can be integrated we get 1 by s minus t. From the first two equations we can solve for t and s in terms of x and y because these are just linear equations very easy to solve and u equal to this 1 by x minus 2y it is defined on the domain whenever x minus 2y is non-zero. So you have to stick to one of them because I do not want my domain x equal to 2y happening. So x is greater than 2y is one option x is less than 2y is another option but when x is greater than 2y I take x positive or x is less than 2y I take x negative why is that? Because only this domain is in contact with the datum curve on which datum curve projection of the datum curve is intersecting only this part or this part it is not intersecting uniformly x bigger than 2y. Now both solutions are local with respect to datum curve solutions. So local solution with respect to datum curve. Observe that base characteristics of the family of straight lines x equal to y plus s see here y equal to t right so y plus s x minus y equal to s that is the family of base characteristics and they fill entire plane still Cauchy problem does not have a global with respect to domain forget about it even with respect to datum curve it does not have a global solution. So this is a manifestation of the non-linearity in the right hand side namely u square in the PDE. Let us look at example 6 here we consider two Cauchy problems for the linear PDE linear PDE minus y u x plus x u y equal to 0 posed of course I do not want coefficients x and y to vanish simultaneously which happens at the origin. So I remove the origin on that domain I consider this equation and the characteristics system of ODE we can directly write down right and base characteristics because of this nature dx by dt is minus y and d by by dt is x so if you compute one more derivative d 2 x by dt square is minus d by by dt that is equal to minus x therefore d 2 x by dt square plus x equal to 0 similarly one can do with y. So solutions of x and y are going to be solutions of y double dash plus y equal to 0 which are combination of cosine and sine and the trajectories will be circles. So base characteristics are the family of circles x square plus y square equal to c square since its positive number we write c square because we do not want to write root c in some other place so we write c square and always make sure make mention this that c is positive so that we do not get confused later. Now the equation for z implies that any solution to the PDE is constant along each of the base characteristics because z is constant right d z by dt is 0 on solutions of this okay so that means on each circle the solution is constant. So if you know at one point on the circle what is the value of the solution then it is the same constant throughout the circle on that circle okay. Cauchy data 1 we consider x equal to s y equal to 0 z equal to s and s r minus 0 okay this is not that what we like but it is okay we continue with this the computation gamma is not a curve obviously it is two pieces okay but never mind adjust for now I guess one can create similar conditions but then they look more complicated than this this is very easy for computation so let me allow me this so I am going to compute with this. So this is the initial conditions and then solutions as I said cost is scientific feature and z is s now from the first two equations for positive s I can eliminate or I can express I think I should not use the word eliminate I can express s and t in terms of x and y I get this and for s negative I get this expression for s that is why negative for s negative positive for s positive t remains same. So in both the cases the function s and t are not defined at points where x is 0 because when x is 0 there is a trouble it is not defined. So since z equal to s the solution is given by u x y equal to s x y therefore u x y equal to this if x is negative x positive now it is defined on r 2 minus x axis and all points of gamma lie on the corresponding integral surface so this is global with respect to datum curve not global with respect to domain because it is not defined everywhere on r 2 minus 0 0 no it is defined r 2 minus x axis. Now another function v defined on entire domain r 2 minus 0 0 given by this formula it is also pd but problem is it is not a solution to the Cauchy problem why for x in r minus 0 v x 0 is root x square that is mod x on one hand but v x 0 has to be x on the other hand because that is the initial condition so both cannot happen particularly for x negative it is not possible so it is not a solution to the Cauchy problem throughout yes if you restrict for x positive sign then yes okay Cauchy data 2 here we look at x equal to s y equal to 0 z equal to h s so it is like correction for the previous thing here mod x is an even function and here given data is not even function that is why there is a problem so now I am going to change it to even function h s where h is an even function still the same problem but again adjust same procedure as above we get u x y equal to a function h of root x square plus y square this is defined whenever x y is different from origin and a smooth function if h is differentiable c 1 function then this is being a composition of c 1 function it will be c 1 function and it will be a solution see now u x 0 is h of mod x and that is equal to h x because it is an even function thus Cauchy problem has a solution defined on r 2 minus origin so it has global with respect to domain solution therefore global with respect to data as well solution is global with respect to both so let us summarize many Cauchy problems are solved using method of characteristics understood the local nature of solutions to first order PDEs this we understood local nature can be in the same two different senses one means with respect to the datum curve another is with respect to the domain of course through two examples we have understood and reasons were different in each of these examples and in a forthcoming lecture an example of a Cauchy problem for Berger's equation will be studied in that example the local nature of a solution arises due to intersecting base characteristics that we will see in a forthcoming lecture so with this we come to the end of a Quasi linear equations and we will then start with a general non-linear equation once again Cauchy problem we will be making a regular comparison to what we did in the Cauchy problem for Quasi linear equations you may say that Quasi linear equation is a special case of a general equation why do two times right why repetition why do not you do the general thing first no because Quasi linear always when you do not understand something you would like to understand with a special case Quasi linear is one such special case where things are easily understood now we try to extend these ideas to the general case that is what is the natural progression in solving problems in mathematics thank you