 In this lecture, we are going to discuss the situation when the transversality condition fails. The outline for the lecture is as follows, first we discuss failure of transversality condition and its consequences, what can be expected in when there is a failure of transversality condition. Then we look at illustrative examples where transversality condition fails, these there are 2 examples, they illustrate the 2 possibilities that we can conclude in the case when transversality condition fails. And then we look back at the 3 examples which were motivating examples which were dealt in lecture 2.1. It was Cauchy problem, 3 Cauchy problem for the same differential equation, same PDE wherein we saw one has exactly one solution, other Cauchy problem has infinitely many solutions and the third one had no solutions. So, we are going to analyze those examples in the light of this failure of transversality condition. In the proof of existence and uniqueness theorem, the transversality condition played an important role in both existence and uniqueness assertions in proving both of them. So, therefore, it is natural to ask what happens if transversality condition is not satisfied. We expect that the Cauchy problem may not have a solution because that is existence is violated, no existence. Second aspect is uniqueness is violated that means it has more than one solution or it may still have one and only one solution, are you surprised? No need to surprise because the theorem if you remember the assumption of transversality condition was only a sufficient condition, under that condition we have proved existence and uniqueness. So, if the sufficient condition is not satisfied, still we may have existence of a unique solution. I would like to recall your attention to one of the very important ordinary differential equation problem. Let me write that equation y dash equal to y sin 1 by y obviously when y is not 0 and 0 when y equal to 0 that is to be expected and then y of 0 equal to 0. So, this initial value problem, okay, the theorem that we have is a Pino's theorem and Cauchy-Lipschitz-Picott's theorem, when we apply Pino's theorem, yes the right hand side is a continuous function therefore there is no problem it has a solution. Now the right hand side is not a Lipschitz function that you can check it. Cauchy-Lipschitz-Picott's theorem cannot be applied and uniqueness be concluded we cannot do that. So, therefore we can expect that maybe there is more than one solution. In ODE's it so happens that when you have two different solutions for the same initial problem there are infinitely many solutions actually so that is the only possibility. So you have no solution or one solution or infinitely many solutions this is the only possibilities for ODE's. Now here it so happens this problem the right hand side is not a Lipschitz function this is not locally Lipschitz. If you call this as f of y there is no y dependent x dependence so I do not write. So this is locally not locally Lipschitz it is not locally Lipschitz. Therefore we cannot apply the theorem but still it so happens this has a unique solution. So that I leave it for you if you are not come across this question already please look at it and show that solution is unique. So violation of sufficient condition means nothing no conclusions can be found that is why you may still have one and only one solution it can happen. So if the transversality condition is not satisfied and a solution to the Cauchy problem is expected that means you the Cauchy problem has a solution then something should happen. The datum curve must necessarily have a characteristic direction at each of its points which means that it is a characteristic curve. Now we have a result lemma 1 it is a very simple result. Let z equal to u x y be an integral surface corresponding to the Cauchy problem for Quasillinear equation QL containing a part of the datum curve which is called gamma dash and that corresponds to you see gamma corresponds to s belongs to y gamma dash corresponds to s belongs to i dash where i dash is a sub interval. Now take a point on gamma dash and denote j 0 s is the precisely the determinant which comes in transversality condition a b here f dash g dash a b is the tangential direction for the base characteristic curve and base characteristic passing through the point f s g s but because it is Quasillinear it is tied up with the value h s also f prime g prime is a tangential direction to the base characteristic curve not sorry not base characteristic curve to projection of gamma which is gamma 2 it is a tangent for that conclusions if j of 0 s naught is 0 that means that determinant is 0 at one point then the datum curve has a characteristic direction at that point what does that mean datum curve has characteristic direction at s naught what is characteristic direction at s naught a p naught b p naught c p naught because p naught is the point f s naught g s naught h s naught and what is the saying that datum curve has a characteristic direction it is tangent which is f prime s naught g prime s naught h prime s naught that triple is proportional to a b c and if it is 0 j 0 s is 0 for every s in a sub interval i then gamma dash is a characteristic curve for the Quasillinear equation proof of 1 so j 0 s naught is given to be 0 right and we want to show that datum curve has a characteristic direction at that point p naught so j 0 s naught is 0 if and only if this is just expansion of that determinant a b one column f prime g prime in the other column that determinant is precisely the left hand side that is equal to 0 since use a solution to Quasillinear equation and h s equal to u of f s g s holes we know this we have let us compute this c p naught g prime minus b p naught h prime what is c p naught from the equation a u x plus b u y equal to c therefore that c p naught is equal to this quantity in the brackets a u x plus b u y at the respective points a and b or a b and u x u y are evaluated into g prime s naught minus b p naught into h prime s naught how do I get h prime s naught from here differentiate this with respect to s take s equal to s naught that will give you this by chain rule u x f prime plus u y g prime now that is equal to after simplification is this but here you observe this thing in the bracket is precisely the left hand side with a minus sign so that is 0 therefore this is 0 so this proves one so one is a we have proved on the last slide that gamma has a characteristic direction at the point p 0 proof of two from one we conclude that gamma dash is a characteristic curve if j 0 s equal to 0 for all s in i prime what is the part one says whenever j 0 s equal to 0 and we have a solution then that has to be characteristic curve that that the the atom curve has characteristic direction at that point but in this case it happens for s in i dash therefore gamma dash is a characteristic curve that is the definition of a characteristic if j 0 s equal to 0 for all s then by lemma gamma itself is a characteristic curve so when such a thing happens the Cauchy problem is called characteristic Cauchy problem the Cauchy problem where the datum curve is a characteristic curve that is called a characteristic Cauchy problem now we have lemma 2 this is what is the result which is what happens when transversality conditions what is to be expected of solutions on how many in number so consider the Cauchy problem for ql such that for every s j is 0 conclusions take a point in gamma if gamma is a characteristic curve then there exists infinite number of integral surfaces infinite number which contain a part of the datum curve gamma containing the point p0 that is always there local solution we have been talking about so this will always be there but here the interesting thing infinite number of integral surfaces if gamma is a characteristic curve earlier lemma 1 said if you have a solution then curve gamma must be characteristic curve if the j is identically equal to 0 now this is opposite kind of converse if j is 0 throughout and gamma is a characteristic curve then not only there is a solution but actually infinite number of solutions exist and if gamma does not have a characteristic direction at any of its points then no solution Cauchy problem does not admit a solution note in this theorem we are assuming that j is 0 throughout for all s in i so how do we show infinitely many solutions exist what we do is this let v denote v1 v2 v3 denote the tangential direction to gamma at the point p0 so imagine this is your gamma this point is p0 this is a tangential direction which is this one now choose any direction w such that the set v1 v2 and w1 w2 this is in R2 is linearly independent in R2 we will see in a moment why this we are assuming this is it possible to choose yes it is possible to choose because once you know v1 v2 that is after all one vector in R2 you can find infinitely many such vectors w1 w2 so that this set is linearly independent and w3 you can simply add to that there is no condition on w3 the only condition is on w1 w2 and v1 v2 so it can be done in fact infinitely many ways you can do this that is the reason why we are going to get infinite solutions we will see that so choose any curve gamma tilde in omega 3 that is we had a gamma we had a point p0 where we had some tangential direction now what we are asking is choose anybody else tangent has to be different something like that this is another curve in fact it can be a curve it can be a straight line as well we are going to see that suppose your curve you take no problem so this is gamma tilde take any curve gamma tilde in omega 3 through the same point p0 such that the tangent to gamma dash at p0 is in the direction of w that one easy way of ensuring that is to take straight line so you have gamma here this is the point p0 so you take some curve no I do not want to take curve so you had a gamma you had a point p0 and here through p0 you fix a direction w and you take a line that is it a line passing through p0 with direction w that is simply given by p0 plus some parameter alpha times w this is a vector form of the line right so of course the tangent will be the direction w so the consequence j gamma tilde p0 will be non-zero because on one hand you had f prime g prime right earlier and here you had that a b because of that you got things 0 now you have taken somebody who is so this was 0 earlier that that means f prime g prime is proportional to a b now I replace this with a new thing w1 w2 this will be non-zero okay because this is not linearly independent sorry w1 w2 v1 v2 are linearly independent therefore as a consequence w1 w2 and f prime is not g prime is not the both these columns are linearly independent so determinant is going to be non-zero now I can apply existence uniqueness theorem so what happened is that initially we had a curve gamma there was some problem because j was 0 we could not apply the existence uniqueness theorem then we chose another curve gamma tilde for which existence theorem can be applied as a consequence we get a kind of surface okay which contains a part of this gamma tilde and as a consequence a part of gamma will also be there because that is that is a characteristic curve so the surface is called s tilde that contains a part of gamma also because gamma is a characteristic curve so tangent plane at p0 2 s tilde contains 2 vectors v and w v is there because the curve gamma is there on the surface and v is a tangential direction basically it is a characteristic direction at p0 and w is also there in the tangent plane so in fact the tangent plane is given by the subspace of r3 spanned by the 2 vectors v and w there is a tangent space and tangent plane will be at the point p0 origin will be at p0 so we have got that so there are infinite number of directions w having the property that v1 v2 w1 w2 is linearly independent we already saw this therefore we get an infinite number of integral surfaces such that a part of gamma lies on each other the part might vary from surface to surface all of these integral surface are pair based distinct because the tangent planes are having different directions okay the orientation of the tangent of course tangent plane is always a two dimensional quantity but it will be changing one guy who is always be there in that direction in the tangent plane is the direction of the characteristic direction but the other direction is the w if you as you change w two planes are not same you have infinitely many solutions because you have infinitely many choices for w so second proof of second one assume that there is a solution then by lemma 1 datum curve must be a characteristic curve and we have assumed it does not it is not a character not only it is not a characteristic curve it has it does not have a characteristic direction at any of its points that is a hypothesis of part 2 of lemma 2 so it contradicts therefore Cauchy problem does not admit a solution so this completes the proof of 2 now we are going to illustrate these two conclusions of lemma 2 with the two examples the first example we are going to consider is this equation x ux plus y uy equal to 2u ux 0 equal to x square is the Cauchy data given for x positive okay let us start solving so first thing is you need to parameterize the Cauchy data which is this x equal to s y equal to 0 z equal to square then we have to write down the characteristic differential equations which is this now we have to solve this with initial conditions so that time t equal to 0 this it passes through points of gamma so this is initial conditions solutions are very simple one can write this equations are linear equations right dx by dt equal to dx so solution is constant times e power t here also constant times e power t here constant times e power 2t what that constant turns out to be coming from this initial data so these are the solutions now what is the next step that the existence uniqueness theorem told us using the first two equations namely for x equal to x ts and y equal to y ts express t as a function of x y and s as a function of x y go and substitute in the third one and propose your solution if all is well that is going to be your solution and here we cannot express explicitly you can see x equal to s e power t y equal to 0 there is no way you can solve for both s and t from this but suddenly we see that we look at the third one that is s square e power 2t that is s e power t is whole square therefore z equal to x square a solution we have an i for a solution maybe and tempted to declare that my solution is going to be u x y equal to x square okay it solves a Cauchy problem of course we can substitute and see it actually solves but who guaranteed that this will be a solution because you have not done the application of the theorem because if you want to apply the theorem you have to check that the transversality condition is met you have not done that so somehow you guessed luckily it is working that is it can you give me one more can you say this is the only solution we do not know okay so this question is still to be settled so let us go go through formally go through the theorem to know that if a solution exists or not we have to rely on the theorem look at the Jacobian that is 0 everywhere so I cannot apply the theorem so anything can happen now can we conclude something from lemma to gamma is a characteristic curve one can check it just means one slightly one more thing you have to check what is the tangential direction to gamma and what is the characteristic direction at this point both will be proportional please check that the lemma to assert the existence of infinitely many solutions done let us find some of them yeah theorem says infinitely many theorem also gives an algorithm which can be implemented we will do that so take a point on gamma say this point p I do not really need to put s naught so I just put s all of us know that s is fixed in this right the point p naught is fixed this is how a typical point looks on gamma s 0 s square fix a w such that w 1 w 2 and 1 0 what is 1 0 it is that v 1 v 2 that is 1 0 in this example that is linearly depend independent done now take gamma tilde as this this if you see just the straight line passing through the point p having the direction w p I have used zeta here w if you write in component form is what you get okay s p is what s 0 s square plus zeta w 1 w 2 w 3 so if you expand you will get this so it is a straight line passing through p having the direction w in fact we may take w equal to 0 1 w 3 okay that means I am setting w 1 equal to 0 w 2 equal to 1 why is that because what you want to show infinitely many solutions even now suppose I show for each such w there is a solution do we still have infinitely many yes because w 3 is still arbitrary w 3 is in R 1 can do much more but I am already showing infinitely many solutions there are many more solutions also solution of the characteristic system of ODE satisfying these initial conditions which are now based on gamma tilde is given by x equal to s e t s e power t y equal to zeta e power t z equal to s square plus zeta w 3 into e power 2 t now we can express t and s in terms of x and y as follows t equal to capital T of x y equal to log x by s and zeta is equal to s y by x please note that s is fixed as far as gamma tilde is concerned substituting in the expression for z of t s we obtain the solution as u tilde of x y equal to x square that is s square into e power 2 t is x square plus s y by x that is the zeta here into w 3 w 3 into e to the power 2 t which is x by s whole square it comes from here so on simplification the solution u tilde become x square plus w 3 x y by s solution to the Cauchy problem with datum curve gamma tilde was obtained as this on the previous slide u tilde of x y is x square plus w 3 x y by s please note this is not one solution this is infinitely many solutions because there is a arbitrary w 3 in the formula so with the choice of w equal to 0 1 w 3 gamma tilde takes the form we introduce gamma tilde with respect to w 1 w 2 w 3 but we made a choice that w will be 0 1 w 3 therefore gamma tilde looks like this x equal to s y equal to zeta z equal to s square plus zeta w 3 zeta belongs to real numbers observe that u tilde of s zeta that is substitute x equal to s and y equal to zeta you get s square plus zeta w 3 but what is s square plus zeta w 3 it is this the z coordinate what is s and j s and zeta their x and y coordinate respectively it means that the entire gamma tilde lies on the surface s tilde does that surprise you because the applying existence and uniqueness theorem gives us only for zeta nearby 0 that gamma tilde lies on s tilde it should not surprise because the existence and uniqueness theorem is a very general one it is applicable for all sorts of Cauchy problems and what we are working with here is a specific Cauchy problem therefore the solution can behave better next question is actually can we apply the theorem of course we have solved and got this expression for u tilde of x y is a theorem itself applicable recall our original problem is a characteristic Cauchy problem where we had this determinant right this s 0 corresponds to a and b and this corresponds to original f prime and g prime and that was 0 for all s so it was a characteristic Cauchy problem then what we did is that this continues to be a 0 now we change to gamma tilde when we change to gamma tilde what is the corresponding f dash g dash that is w 1 w 2 which is 0 1 which is equal to s and which is not equal to 0 because we are in the region s greater than 0 so the theorem is applicable and that will give you as you know local solutions with respect to the datum curve but however we have obtained here by the computations we see that we indeed have a global solution with respect to the datum curve. Observe that u tilde of x 0 equal to x square also holds what does that mean that means that entire gamma lies on s tilde that means our original datum curve itself lies on s tilde so not all these solutions are also global with respect to domain they are in fact defined on R 2 okay they are defined on R 2 so there are infinitely many solutions. Now let us look at the second example this I think we already solved using Lagrange's method u x plus u y equal to 1 if not let us do again let us do now so the characteristic system of ODE is dx by z equal to dy by 1 equal to dz by 1 so integrating this set of equations we get this x minus z square by 2 equal to c 1 the other one will give you y minus z equal to c 2 therefore Lagrange method says f of c 1 c 2 equal to 0 that means take any arbitrary function f and this equal to 0 of course it remains that this is valid for only those x y z for which this kind of quantity this tuple lies in the domain of f that is always there second thing is that you should be able to solve for z in terms of x and y only then I would like to call this as a solution okay that is done now we will substitute c 1 c 2 so now I think we will find the f using the Cauchy data that should fix the f then we will get the solution that is the idea whether we will be successful or not we will see so x equal to s square y equal to 2 as z equal to s this quantity equal to c 1 substitute for x equal to s square and z is equal to s you get this so that will give you s square equal to 2 c 1 now from the other one y minus z equal to c 2 when you substitute the values for y and z you get s equal to c 2 therefore there is a relation between c 2 and c 1 that we get c 2 square equal to 2 c 1 so we are going to substitute for c 2 whatever that y minus z and c 1 this that gives a quadratic equation for z when solved we get this expression y plus r minus root 4 x minus y square by 2 now gamma 2 the projection of datum curve lies on this curve of 4 x minus 4 x equal to y square therefore this function is not differentiable because whenever 4 x minus y square is 0 there is trouble square root function is not differentiable so u is not differentiable at any point of gamma 2 therefore Cauchy problem has no solution whatever we have obtained now using method of characteristics we will attempt to solve the same problem again so we have to write the parametric form of gamma done then system of characteristic ODE is this now we have to solve this with initial data this so that at t equal to 0 the characteristic curve lies on the datum curve done when we solve this is the expression we get for x y and z so to know if a solution exists we have to rely on the existence and uniqueness theorem because it is not obvious whether I can I can use the equation for x and y and get an expression for t and s so let me do that Jacobian is this it is 0 so Jacobian is 0 therefore existence theorem cannot be applied now what does lemma to ask it asks whether it is a characteristic curve if it is a characteristic curve answer is infinitely many solutions so we check what is it is it a characteristic curve or not the characteristic direction at any point on gamma is the a b c that is s 1 1 but tangential direction is 2 s 2 comma 1 so both are not proportional they are not parallel okay so therefore therefore gamma is not a characteristic curve therefore the second part says you do not have solution conclusion 2 of lemma 2 says it has no solution okay now let us go back to the three examples that we were discussing in lecture 2.1 that is this recall the PDE was u x equal to C u in fact it is like ODE there is no t dependence and then Cauchy data 1 2 3 Cauchy data 1 had unique solution note where is it prescribed it is prescribed on the t axis while Cauchy data for 2 and 3 are prescribed on the x axis in one case you have infinitely many solutions in the other case there are no solutions then we ask this question why they behave differently PDEs are same and on t axis you have uniqueness on x axis you may have many or no solutions then we had this question who is special is a t axis or x axis and we said answers later now it is a time for answers Cauchy problem 1 j 0 s so we are trying to use lemma 2 and try to answer j 0 s is 1 therefore we have existence uniqueness theorem therefore uniqueness is what you expect done it is in tune with the assertion of existence uniqueness theorem. Now second Cauchy problem we had j 0 now the question is is it a characteristic curve or not so transversality condition is not satisfied fine at every point on gamma the characteristic direction is f prime g prime h prime which is equal to 1 0 c into e power c s and it coincides with the tangential direction therefore it is a characteristic curve therefore there are infinitely many solutions that is what lemma 2 also says Cauchy problem 3 j is 0 so we ask whether the curve gamma is a characteristic curve or not it is not a characteristic curve so the conclusion 2 of lemma 2 says no solution so that is the thing here transversality condition is satisfied so we do not go further here we check gamma is a characteristic curve because a b c is this f prime g prime h prime is this and here gamma does not have characteristic direction anywhere therefore no solutions this is a characteristic Cauchy problem we have infinitely many solutions the transversality transversality condition is not satisfied for both Cauchy problems 2 and 3 datum curve turned out to be a characteristic curve in the case of Cauchy problem 2 that is why we had infinitely many solutions datum curve is not a characteristic curve for this Cauchy problem 3 therefore no solution now let us answer the other part which x is special t or x and y special things have to be handled carefully so you answer for yourself for this question because the answer lies in special things have to be handled carefully something is not special does not matter you give anything you want know there is no problem so existence uniqueness theorem asserted existence of a unique solution for Cauchy problem if transversality condition is satisfied denote the corresponding Jacobian as J0 S0 if this is non-zero then it will be non-zero for S belonging to an interval containing S0 for the reasons that things are involved in the determinant or continuous functions of S and determinant itself is a continuous function so given the local nature of the assertions in existence and uniqueness theorem the following cases exhaust all the possibilities in when the transversality condition fails at S0 J vanishes at all points of an interval of some interval containing S0 nearby S0 is what we are worried about and some interval J is 0 or it does not vanish at any point of some interval containing S0 except at S0 that is the isolated point in some interval in rest of the points of the interval J is not vanishing but only at S0 it vanishes or it can happen like this kind of limit point you find a sequence of Sn so as the J is 0 at the point J0 Sn and Sn goes to S0 otherwise it is not 0 imagine a function like X sign 1 by X 0 is the limit point of zeros of this function something like that now lemma 2 answered the cases 1 and when the case 1 happens when J is identically equal to 0 on some interval containing S0 and theory seems to be elusive when situation 2 or 3 occurs one has to look from case to case and make conjectures and I do not think there is any complete theory available written in a textbook. So let us summarize we have learned how to solve a Cauchy problem using method of characteristics we have understood what may happen if transversality condition is not satisfied and in the next lecture we solve some more problems from first order partial differential equations up to quasi-linear equations and then we will move on to general non-linear equations or general equations later on thank you.