 In this lecture we are going to discuss about certain curves of discontinuity associated to second order quasi linear partial differential equations. So we start with a brief review of the lecture 3.1 and then we move on to discuss curves of discontinuity. So as you know the second order quasi linear equations we are denoting by 2ql and it stands for AUXX plus 2BUXY plus CUYY plus D equal to 0 where A, B, C, D are functions of this X, Y, U, UX and UY. So they are defined on some subset of R5 which is called omega 5 and a special case would be a linear equation second order linear equation it has the general linear equation looks like this. So let us start with the review of the previous lecture. We have considered the following Cauchy problem in the lecture 3.1. What is that? Find a solution to the second order quasi linear equation satisfying U of FSGS equal to HS and the normal derivative of U at point FSGS is chi S. What is FSGS? It is a curve parametrically given in a plane gamma 2 and the curve is a regular curve that means F dash and G dash do not vanish simultaneously. So along this curve we are specifying the value of U as H and the normal derivative of U as chi. So these functions H chi are given. So this is often called Cauchy data. Cauchy data along gamma 2. So we observed that the first order derivatives of U at points of gamma 2 can be computed using the Cauchy data. For that what all was needed is that the curve gamma 2 is a regular curve that was good enough. Then we also try to compute second order partial derivatives of U again along the curve gamma 2. It can be computed using Cauchy data and now we have to involve the partial differential equation as well. So the second order quasi linear equation using these two we can solve but one more condition was to be met. So that was delta of S is not equal to 0. What is delta of S? It is this expression. C F dash square minus 2 B F dash G dash plus A G dash square we know what F and G are. These are the functions which describe the curve gamma 2 parametrically. Then what is zeta S? It is a 5 to pull which is F, S, G, S, H, S, P, S, Q, S. We have already solved for P, S and Q, S using chi and H. Therefore this makes sense and we can ask this condition. If this is non-zero delta S not 0 then we can compute second order derivatives and we have also shown that higher order derivatives can also be computed. No further extra conditions are needed. Now we would like to understand this condition in more detail. What is this curve? What is this? Where delta S is identically equal to 0, how that curve looks like? So let us assume to start with that the gamma 2 that we considered is not parametrically defined but let us say it is a graph of a function of the type x equal to phi y. It is a graph of a function of the variable y, x equal to phi y. Then what happens to gamma 2? x equal to F, S that will be phi S, y equal to G, S which is S. Y we parameterize as S, then x becomes phi S. So this is a parametric representation and S belongs to that set where y belongs to wherever this function phi is given the domain S belongs to the same domain. We are not writing that right now. Now what will happen to this equation? Now we need to substitute what is H S, P S and Q S. H S is still general. P S and Q S are to be determined from H S and chi S. So these are still unknown but what is known is F dash. F dash square is phi dash square, G dash is 1. So this equation becomes this equation. Of course if you want to know what function satisfy this condition we need to still know this H, P and Q. That is the problem with this equation because the equation is quasi-linear equation. So for a quasi-linear equation if you are interested in this question then you must be given a solution of the equation. Then you can ask what is that curve which will have this property. That can be now determined because solution is given therefore you know what is P and Q. So it makes sense. On the other hand if the equation is actually linear the A, B, C are not functions of zeta S at all. Zeta S is a phi tuple. They are simply functions of the first two variables F S and G S. And what is F S and G S? That is phi S and S. So therefore you know explicitly C, B and A only in terms of phi. So it will be a differential equation for phi. So it further reduces in the case of a second order linear equation to a simpler equation. For a quasi-linear equation however we need to be given a particular solution for which we can ask what are the curves x equal to phi y along which there is a problem in finding higher order derivatives. Problem in the sense we are unable to our scheme of things fails. Our scheme to compute the second order derivatives onwards fails because delta S is 0 there. So as we saw these are special to the PDE because the definition of course involves the equation. So for the linear equation this gets simplified to this x equal to phi y. So this is a differential equation nonlinear first order degree 2 nonlinear and coefficients will involve only two variables now because we have linear equation. So one can hope to solve this for phi. Then we would have got the curve x equal to phi y where phi is a solution of this ODE is a curve along which delta S is identically equal to 0 or there are troubles in solving for higher order derivatives. In other words along these curves there is some problem to determine higher order derivatives of the solution or of a possible solution. So even if you know solution there is a problem to determine that because of delta S being 0. Now the curves comma 2 for which delta S identically equal to 0 holds are special to the PDE as PDE also plays a role in their definition apart from gamma to itself. The same curves appear in a different context also. So for piecewise smooth solutions of the second order quasi-linear equations the curves of discontinuity okay curves of discontinuity also satisfy delta S identically equal to 0. So let us we are going to state a result in the form of a result it will be easier to remember. So hypothesis and notations suppose you have an open and connected set in R2 and take a curve in R2 gamma that divides omega into two parts. So that means that we have this let us say this is omega and we have a curve gamma which cuts this into two pieces. One is omega 1 other one is omega 2. So what is omega now? Omega consists of omega 1 omega 2 and this curve gamma the part of the curve gamma. We are going to assume that the curve gamma is given by x equal to phi y. We have already seen its interpretation in lecture 3.1 why do we consider x equal to phi y? Once again we will give at the end of this result. So given V1 in C of omega 1 bar which means that you are given a function V1 here which is continuous on omega 1 of course omega 1 open set but it should be continuous up to the boundary of omega 1. In other words what actually we want is its meaning on this curve. So therefore V1 on the curve gamma makes sense. Similarly V2 is in C of omega 2 bar means that the values of V2 on the curve gamma makes sense. It will guarantee that apart from of course on this boundary also they make sense because it is continuous up to the closure of the domains in this case omega 1 in this case omega 2. If such functions are given we can look at the value of V2 on gamma and V1 on gamma it makes sense meaningful therefore we can look at the difference. Let us see. Let us define a function V in omega 1 it is V1 in omega 2 it is V2. Let box of V denote the jump in the values of V across gamma. That means the definition is take a point x y take a point x y on gamma is omega 1 this is where V1 is defined this is where V2 is defined. We have seen V2 of x y makes sense when x y is in gamma let us call this part as the gamma may not be the outside part okay. V2 of x y makes sense V1 of x y makes sense because V1 is also continuous up to the boundary there in particulars values on gamma make sense and we can consider the difference. You could have even taken V1 minus V2 there is no problem and we are considering V2 minus V1. So this is called jump in the function V that means you have a function V defined on omega 1 here you have a function here you have a function which is continuous so that the values make sense along this curve gamma then you look at the jump in V as V2 minus V1 at points of gamma. Let U, Ux and Uy be continuous on omega and the restrictions of U to omega i they are C2 functions because we want them to be solutions to the PDE so that is why the C2ness and we want omega i bar because we are going to consider the jumps in second order derivatives. First of all derivative should be meaningful on the curve gamma okay and this curve gamma. So if I assume U is C2 of omega 1 bar we may put 1 on the head U1 is C2 of omega 1 bar it means that second order derivatives of U has are defined on gamma because they are continuous in omega 1 bar that is the reason why we have this condition. So let gamma be given by x equal to phi y where phi is a C1 function. Now the jumps in Uxx, Uxy and Uyy are not independent of one another okay it means they are interrelated to start with it looks like yes Uxx is jump in second order Uxx derivative this is jump in xy derivative this is jump in yy derivative why should they be connected. So let us denote lambda y as a jump in Uxx at a point and gamma a point and gamma looks like phi y, y so then this is the result jump in Uxy is in terms of lambda y and phi dash and jump in Uyy is lambda into phi dash square we are going to prove this so they are related observe that if lambda y is 0 what will happen lambda y is 0 means the right hands are 0 here which means jump in Uxy and Uyy are 0s that means if jump in Uxx is 0 then jump in Uxy as well as Uyy are 0 okay let us prove the conclusion 1 the restrictions of a function defined an omega to the regions omega 1, omega 2 are denoted using superscripts that is we had this as the omega and we had a curve which is making it into 2 parts omega 1 and omega 2 and suppose we have a function U defined on omega or any function U, Uxy we denote the restriction of U to omega 1 by U1 and here by U2 so Ui is U restricted to omega i. Since U, Uxy are assumed to be continuous on omega these functions are continuous across gamma also in other words there are no jumps across gamma so jump in U is U2 minus U1 phi y, is a point on the gamma similarly the jump in Ux defined by Ux2 minus Ux1 that is also 0 jump in Uy is also 0 so on differentiating the equations 2b and 2c the last 2 equations differentiate them because differentiating the first equation will not give you anything because if you differentiate this with respect to y what will you get nothing useful because we want to now consider jumps in second order derivatives that is the reason so we differentiate the last 2 equations on the previous slide we get this by differentiating the first equation and this by differentiating the second equation. Now in terms of the jumps this equation is nothing but jump in Uxx into phi dash plus jump in Uxy equal to 0 that is the first equation and the second equation is jump in Uxy which is this into phi dash and jump in Uyy equal to 0. So conclusion 1 of the theorem follows from the last 2 equations because we called Uxx jump as lambda then jump in Uxy is minus lambda phi dash therefore jump in Uxy is minus lambda phi dash that will give you jump in Uyy is minus lambda phi dash into phi dash that is the conclusion 1. Let us look at the conclusion 2 of the theorem for that we need to assume the hypothesis are U1 and U2 solve the quasi-linear equations in the regions omega 1 and omega 2 respectively. Let y be such that lambda y is not equal to 0 recall lambda is the jump in Uxx at the point phi y comma y conclusion denoting by zeta y equal to this quantity phi y comma y U at the point phi y comma y Ux at the point phi y comma y Uy at the point phi y comma y recall the point phi y comma y means x equal to phi y this is how a point on gamma looks like. So the following equation holds we are going to show this equation holds whenever lambda y is not equal to 0 at that point this equation holds. So if you assume lambda y is not equal to 0 at all the points then the equation holds for all y then this equation is same as delta s identically equal to 0. Proof is very simple we are going to assume that U1 and U2 are solutions therefore we can write down these equations in omega 1 and omega 2 with appropriate superscripts. So we have this now we have two equations we have to subtract one from the other and we have assumed that there are no jumps in Ux and Uy therefore the coefficients a, b, c, d they all are like this they are functions of x, y, u, u, x, y there is no jump in them so this a will come out to be a. So only you pick up the jump in Uxx so 2b will remain as it is because it is a continuous function and it is the same in both therefore along gamma what I mean by same in both is this. You have an expression for let us say here a is x, y, u1 of course of x, y, ux1, x, y, ux, uy1, x, y we have this and here what we have is exactly same but with 2. Now on this curve if you x, y is here then both are same because there is no jump Ux2 is same as Ux1 when x, y is on this curve gamma that is why when you subtract they come out as it as they are and the d term gets cancelled because the same in both of them when you are looking at points of gamma. So you pick up only the jumps jumps in Uxx jump in Uxy jump in Uyy they come here. So we have this. So using the conclusion one of theorem in the last equation we get 0 equal to a of zeta y lambda y minus 2b of zeta y lambda y phi dash plus c of zeta y lambda y phi dash y square. So if you take lambda y common what you get is this a minus 2b phi dash plus c phi dash square assuming that lambda y is not equal to 0 along gamma we get this the one in the brackets must be 0. If lambda y is not equal to 0 for all y then this the one in the parenthesis must be 0. So we get this different equation and this equation is the same as delta s equal to 0 for s in i same equation when x equal to phi y. So we have assumed that the curve is of the form x equal to phi y. The conclusion three of the theorem is as follows hypothesis assume U1 and U2 solve the linear equation in the regions omega 1 and omega 2 respectively the linear equation and U possesses third order partial derivatives that is required for stating this conclusion and have jump discontinuities across gamma. Let y be such that lambda y is not equal to 0 then lambda satisfies the following ODE 0 equal to 2 into b minus c phi dash into lambda dash plus this quantity into lambda. So this is a linear ODE with variable coefficients because they depend on y this depends on y. So it is a linear ODE with variable coefficients of course if b minus c phi dash equal to 0 then this will be a singular ODE otherwise it is a linear ODE. Now observe that if b minus c phi dash is not equal to 0 and abc phi are smooth functions then lambda y equal to 0 at isolated points implies that lambda y is identically equal to 0. Let us understand this carefully if lambda y is equal to 0 at some point this ODE may not hold but we are here the hypothesis lambda y equal to 0 at isolated points. That means the points where lambda y equal to 0 can be reached by the points where lambda y is not equal to 0 at which the ODE holds this ODE makes sense because whenever lambda is not 0 at some point the ODE happens the lambda satisfies this ODE. And you see this is equality as people say equality is a closed condition therefore it follows that even if lambda y equal to 0 at isolated points this ODE continues to hold and if you are assuming all these functions are smooth and this is non-zero then it turns out that the quantity in these brackets and the quantity in these brackets are smooth functions and they are locally lipsious functions then we can apply uniqueness theorems for the initial value problems and conclude that lambda y is identically equal to 0. So lambda y by definition is jump in u x x which is this differentiate both sides of this equation you get lambda dash with respect to y therefore you differentiate this with respect to x you get u triple x but phi is there therefore phi dash and from here you get u triple x into phi dash now differentiate u x x with respect to y variable you get this similarly this you get this. Therefore, we can express d lambda by dy as jump in u triple x into phi dash plus jump in u x x y now similarly we can also get derivative of u x y you write the equation for u x y jump differentiate with respect to y you will end up with this relation. Now we know that u x y jump is minus lambda times phi dash that is a conclusion 1 therefore this equation I can write it as this. Now we are going to use u i solve the linear equation in the domain omega i so we have this 2 equations 1 in omega 1 1 in omega 2 along gamma we will consider because we are going to take jumps. So first is to differentiate this with respect to x because we are interested in third order derivatives differentiate with respect to x then subtract one from another we get this expression. Now once again use the conclusion 1 of the theorem which expressed basically the relations of jump in u x x u x y and u y y in terms of lambda which is actually jump in u x x so we get this relation. So we have 3 equations this this and the one we just obtained this. Now eliminating the jumps in the third order derivatives of u across gamma and using this relation that a minus 2 b phi dash plus c phi dash square is 0 that is the equation phi has to satisfy right even for quasi linear equation therefore for linear equation also. So this equation is satisfied so how do you eliminate that is what you have to look at. So we want to remove these conditions we have 3 relations in them may be solve for them that is what it is and substitute and we get the ODE in conclusion 3 so this is left as exercise to you this is a simple algebraic exercise. Now we may interpret this lambda as the intensity of the jump in u x x intensity of the jump in u x x across gamma this interpretation is taken from the book of Fridgen on PDEs. So if lambda y not is 0 for some y not that is lambda 0 at some point and if the initial value problem that we had for the ODE has a unique solution what is the ODE that we were considering it is a linear equation for lambda which we saw 2 into b minus c phi dash into lambda dash plus a huge expression into lambda. If that has a unique solution of course it is a homogeneous linear equation 0 is always a solution and if it has uniqueness also then it must be that lambda is identically equal to 0 we already did this conclusion. So if u x x is continuous at some point this is the interpretation in terms of u x x what is lambda after all it is a jump in u x x if jump is 0 means what u x is continuous at that point then it is continuous at all points of gamma and therefore once you have u x is jump is 0 jump in u x y and u y y is also 0 from conclusion 1. So location and speed of jumps if the variable y is given the interpretation of time the equation x equal to phi y gives the location of the jump in u x s for various time instances at different instances of time the speed of propagation of discontinuities is given by dx by dy which is equal to phi dash of y dx by dy is phi dash of y and satisfies the differential equation a of zeta y minus 2 b zeta y phi dash plus c of zeta y phi dash square where zeta y is given by this phi tuple. So let us summarize this equation which is the ODE which is there on this slide it appeared in two different contexts in lecture 3.1 how it appeared if the Cauchy data is prescribed along the curve gamma 2 x equal to phi y where phi does not satisfy the above ODE derivatives of all orders for a solution to 2ql can be determined at all points of gamma 2. In this lecture if gamma given by x equal to phi y is a curve of discontinuity for second order partial derivatives for a piecewise smooth solution of 2ql as in the theorem that we have stated today in this lecture then phi solves the above ODE. Important question is for a second order quasi linear equation how useful is this equation since to write down this equation zeta of y it requires the knowledge of a solution to the 2ql that needs to be given right. So how is it helpful in identifying these curves x equal to phi y where phi solves this equation. Once the function is known can we find cause of discontinuity directly do we still need to solve this ODE answers if a solution is given then in principle possible curves gamma across which second order derivatives may have jump are already known. However the ODE gives an analytic characterization of such curves and thus may still be useful. Of course for semi linear equations the ODE is useful in both the contexts determine curves along which a Taylor series for a solution may be obtained or identifying curves across which piecewise smooth solutions will have discontinuities in the second order derivatives. So in forthcoming lectures the ODE will play an important role in reducing a linear second order PDE to a simpler form. In the simpler form the coefficients of second order derivatives will be constants to start with in a second order linear equation coefficients are functions of x and y. But this ODE will help us in obtaining a simpler form of these equations in which the coefficients of the second order derivatives are constants. The simpler form would be obtained using change of coordinates. The ODE plays an important role in determining the change of coordinates. Thank you.