 We are going to begin the study of second order partial differential equations. In this lecture we are going to study about special curves associated to second order partial differential equation. So the outline for today is we start with some illustrative examples exactly like how we started off our study of first order partial differential equations where in lecture 2.1 we looked at 3 Cauchy problems which exhibited all the 3 possibilities for the number of solutions. Unique solution when the data is datum curve is of certain type and when it is of another type it was either 0 solutions or infinitely many solutions. Then we will make an attempt to solve a Cauchy problem for a second order quasi linear PDE. So second order quasi linear PDE in 2 independent variables the most general such equation is of this form AUXX plus 2BUXY plus CUYY plus D equal to 0 where ABCD are functions of 5 variables. So they are defined on open subset omega 5 of R5. So in the above equation we refer to that as 2 QL second order quasi linear equation we suppress the dependence of A, B, C, D on X, Y, U, X, U, Y. Otherwise the equation will be very long when it is understood there is no need to repeat it. So that is why the dependence is suppressed here. So A stands for A of X, Y, U, X, U, Y etc. Now second order linear PDE we will consider this also in this chapter. In fact the second part of this chapter will be as exclusively dealing with second order linear PDE and a general such general form of such equation is here. So we refer to that as 2L second order linear partial differential equation. So let us look at some illustrative examples. Let us start with this second order partial differential equation UXX equal to 0. I have to mention it is in 2 independent variables because it is not clear from the equation. So it is general solution we can write down. This is actually a ODE with respect to X. Therefore when you integrate first time UX of XY will be constant with respect to X. So it will be some arbitrary function of Y that is K1Y. Integrate once more you get UXY equal to K1YX plus K2Y. See solutions of UXX equal to 0 or straight lines if it is a ODE straight line will have A they look like AX plus B. Now because this there is a Y variable involved A and B will be functions of Y also. Now what are K1 and K2? K2 of Y if I want to find out I need to finish this term kill this term. So I put X equal to 0 that means U of 0 Y this term is gone what remains is K2Y. So K2Y is nothing but U of 0 Y and what is K1Y that looks obvious you are differentiate with respect to X and UX of XY in fact is K1Y but I am taking X equal to 0 to be uniform with this because later on we are going to look at prescribing conditions to solve UXX equal to 0 conditions like our Cauchy data that we have seen in first order PDE. Therefore let it be U0Y and UX0Y then what we get is K1Y and K2Y. So therefore this analysis shows that you have complete freedom to prescribe U of 0 Y and UX of 0 Y if you want to solve UXX equal to 0. Now let us look go to the second example equation is the same we are not changing the equation UXX equal to 0 therefore the general solution continues to be the same. Now suppose we want to prescribe U of X0 arbitrarily what should be that can we do it is that allowed by the equation because this equation immediately the solution is coming like this right. So therefore asking this question is same as asking whether equation allows us. So what is UX0 when I put UX0 Y equal to 0 I get K1 0 times X plus K2 0. So it is not allowed by the equation you cannot have arbitrary functions UX0 must look like AX plus B for some real numbers A and B. Let us prescribe UX0 equal to AX plus B prescribing UX0 equal to AX plus B means that K1 0 is A K2 0 is B we just saw that on the previous slide. So since UX0 equal to AX plus B the derivative of U with respect to X is determined on the X axis UXX0 will turn out to be A. So there is no way that we can prescribe UX on the line X0 it is not possible. Then we ask the question can we prescribe UI in that case UI of X comma 0 let us find out. So UI of X0 from this general solution will look like this now derivative with respect to Y that means K1 dash Y X plus K2 dash Y when Y equal to 0 it is K1 dash 0 times X plus K2 dash 0. This means even UI of X comma 0 must be like CX plus D it must be linear function for some constant CD in R. So therefore to conclude we have the following two scenarios when UX0 and UIX0 are prescribed what are they this is just to recall that the general solution is this for UXX equal to 0 what are the two scenarios prescribe UX0 equal to AX plus B and UIX0 equal to CX plus D. Now prescribing UX0 equal to AX plus B fixes the values of K1 and K2 at the points at the point 0 X equal to 0 K1 0 is A and K2 0 is B. Now the other condition UI of X0 equal to CX plus D that fixes K1 dash 0 equal to C K2 dash 0 equal to D. Therefore there are infinitely many solutions to the Cauchy problem because these two conditions namely UX0 equal to AX plus B and UIX0 equal to CX plus D does not determine both K1 and K2 the functions what are all the things which are determined by this condition are simply the values of the function and the derivative at the point 0 for both K1 and K2. So you have so many functions which satisfy this criteria that is why you have infinitely many solutions. Now the second scenario is at least one of them you have not prescribed as a linear function what will happen no solutions no solution to the Cauchy problem. So if you compare both the examples PDE is the same in both the examples namely UXX equal to 0 when Cauchy data is given on Y axis solution is unique when Cauchy data is given on X axis two possibilities exist infinitely many solutions are no solutions. So recall the three Cauchy problem that we considered in lecture 2.1 for first order PDEs where we had similar observations the curves gamma 2 which give rise to 0 or infinite number of solutions turned out to be a special curves for the PDE and these curves were called base characteristic curves later on. Now the natural questions are are they a special curves for every second order PDE how many will be there how to find them. Next few lectures are devoted to finding answers to these questions. So solving Cauchy problems for second order PDEs we cover some preliminaries. So we are going to pose a Cauchy problem and then implement a classical strategy to solve that before posing Cauchy problem we need to introduce a few terminology we are going to do that. So let gamma 2 denote a planar curve described parametrically by gamma 2 X equal to FS Y equal to GS SNI where I is an interval in R and FGR C1 functions. Further assume that gamma 2 is a regular curve what does that mean for every SNI F dash G dash is not equal to 0 0 and geometrically speaking gamma 2 possesses a well defined tangent at each of its points. We have come across the notion of a regular curve even in the context of first order PDEs. Let N of FS GS denote the unit normal to gamma 2 at the point FS GS in gamma 2. FS GS is a point on gamma 2 and N denotes the unit normal to gamma 2 at that point. Defined by this because unit normal is not unique it will be there will be two choices for example if this is your gamma 2 this is the tangential direction and what you have here is the normal direction. So plus or minus of each other we do not care which one we are taking for this problem it should be given by this we are giving the formula here N FS GS equal to this. Notice this is well defined if the denominator is not 0 that is precisely the assumption of the regular curve. So we are now ready to state the Cauchy problem given two functions we will see where they appear H and chi. Cauchy problem for the second order Quasillinear equation consists of finding a C2 function that is twice continuously differential function such that U is a solution to the PDE which is a second order Quasillinear equation 2ql and U satisfies the two conditions U of FS GS equal to HS that means U is prescribed on points of gamma 2 as H of S and dou U by dou N which is called the normal derivative of U is also prescribed as chi S at every point of gamma 2. So suppose this is your gamma 2 and you take a point here at this point this point is like FS GS this is how points look on gamma 2 you are prescribing what should be the value of U at this point and you are also prescribing dou U by dou N at this point the normal derivative why not any other derivative that question we will discuss at the end of this lecture why not any other derivative. So actually if you see the normal direction is like that right or maybe any of the direction let us take this side you can actually define the directional derivative in any direction that can be prescribed. So what can be prescribed is dou U by dou V that can be prescribed that is also fine what all you should not prescribe is the in the direction of the tangent you should not prescribe if you recall if you consider U x 0 equal to some function H x U x is already determined okay. So therefore you cannot prescribe this with freedom and what is U x x 0 it is a directional derivative of U in the direction 1 comma 0 that is the direction of the x axis. So that is what is called a tangential derivative okay and at every point the direction of the tangent and the normal they will be linearly independent. So you can prescribe 2 derivatives but one derivative tangential derivative is already determined if you have prescribed the function U therefore there is U this you cannot prescribe therefore any other directional derivative we can prescribe where V is independent of the direction 1 0 but to be very clear we are prescribing on a normal because tangent and a direction which is immediately connected to a tangential direction is a direction perpendicular to that which is a normal direction. So that is why we prescribe dou U by dou N that is a secret you can prescribe any other derivative also and we require as usual the condition should be met for S belonging to a sub interval of phi which means we are looking at local with respect to data kind of solution. So geometrically speaking if you define a space curve gamma in R 3 by putting z equal to H s we get gamma 2 will be the projection of gamma 2 xy plane and the surface z equal to u x y defined by solution or the Cauchy problem will contain a part of this gamma okay. Now we discuss a classical strategy to solve the Cauchy problem. The goal is to find solution to Cauchy problem near points of gamma 2 because functions will be defined in a neighborhood of gamma 2 the same thing was true for the first order partial differential equations also. So take a point on gamma 2 p0 in terms of x and y you may call x0 y0 in terms of the parameter running on the gamma 2 fs0 gs0. Determine derivatives of all orders of a possible solution at p0. Determine derivatives of all orders propose a Taylor series around the point p0 using the information on derivatives at p0. What information do we need to propose a Taylor series of a function all the derivatives at a particular point in this case p0 that you have determined in the this step determine all the derivatives. So Taylor series can be proposed and hoping that the series converges and it would be a solution to the Cauchy problem that is the hope one needs to prove that. This is the essential idea behind the proof of Cauchy-Kowalski theorem for details you may consult the book of partial differential equations by Fritz John you will find the details there. So in other words somebody comes to you and tells you that boss I know that this Cauchy problem has a solution which can be expressed in Taylor series format. It has a Taylor series expansion in other words he is telling you that solution is real analytic. Now your job is only to find that to find that what all you need to do is find all partial derivatives of the function at the point p0. If you can determine them uniquely then you can't hold off all the derivatives and propose that series Taylor series. And since somebody told you that he has a Taylor series expansion you hope that this will be solution to that. But to implement this strategy we would require the data in the problem namely the A, B, C, D to be smooth to be as many times differentiable as we want. Similarly F, G and H which are prescribed functions or maybe H and Chi yeah F, G are determined defined by gamma 2 gamma 2 is defined by F, G and then we are given the Cauchy data in terms of H and Chi. So all of this of course we need to assume are C infinity functions then only we can implement this strategy. We limit ourselves to just computation of all derivatives we are going to inquire into the possibilities of computing all derivatives at a point p0. Can we do it or not? Whether somebody stops us from doing that if so who is that we will identify. Of course we have no a priori knowledge of the solution the person who told there may be a solution he is not given as the formula so that I can compute derivative using the function. No it is not that you are given a function and then find its Taylor series it is not the case. You are thinking that there is a solution which has a Taylor series expansion and you are trying to find out if such is the case what are the derivatives and what is available to you is only the Cauchy data and the PDE these are the only two things that you can use nothing else. So we do not discuss the convergence aspects of the formal Taylor series which needs to be proposed after computing the partial derivatives of all orders. So for implementing this strategy we need to assume that all the functions involved in the Cauchy problem namely a, b, c, d, f, g, h, chi are C infinity that means they have derivatives of all orders. Now we are going to drop the subscript 0 in S0 and we write S with understanding that it is fixed but otherwise arbitrary in I. Now let us see the computation of first order derivatives using only the Cauchy data and the PDE and for brevity in notations let us introduce ux at a point fsgs on gamma 2 as ps similarly uy at a point fsgs as qs. So these functions are defined on gamma 2 we are just introducing we do not know ux uy yet we need to determine ux uy. So using these notations the normal derivative condition takes this form minus pg prime plus qf prime by root f tax square plus zeta square equal to Cauchy and dou u by dou n is gradient u dot n gradient u is ux and uy dot n is minus g prime f prime divided by root f dash square plus g dash square that is why we get this equation solution must satisfy hs equal to u of fsgs u of fsgs also right. So let us differentiate this equation because we want to get an equation for ps and qs. So we differentiate this apply chain rule so from here we get this but ux is ps uy is qs therefore that is ps f dash plus qs g dash s. So we have got one more equation. So we had one equation on the previous slide and one more equation on this slide both of them are linear with respect to ps qs. So let us recall both the equations in one place look we want to determine ps and qs this is one linear equation featuring ps qs f prime g prime h prime are known here also g prime f prime and chi are known so this is also a known linear equation in p and q. The coefficient matrix is invertible what is the coefficient matrix the first equation is f prime of s g prime of s second one is minus g prime of s divided by square root of f prime square plus g prime square and here it is f prime by square root of f prime square plus g prime square into ps qs this is a system equal to h prime into chi not chi into h prime and chi. Now what is the determinant of this it is f prime square plus g prime square divided by square root of f prime square g prime s square therefore determinant is equal to square root of f prime square plus g prime square and that is not equal to 0 due to the regularity of the curve so therefore there is exactly one solution for ps and qs so we can find ps and qs uniquely. So both the first order partial derivatives have been determined at all points of gamma 2 with this in fact we are interested in determining at some point of gamma 2 that we have fixed since the point is arbitrary we are saying at any point in gamma 2 using only the Cauchy data the PDE did not play any role so a remark computation of first order derivatives at points of gamma 2 required the knowledge of u and its normal derivative on gamma 2 only these not surprising the Cauchy data contains information on directional derivatives of u in two independent directions what are they they are tangential through u of fs gs equal to hs normal through the normal derivative which is given explicitly. Since information on tangential derivative is inbuilt in this condition u of fs gs equal to hs one needed to prescribe derivative in any other direction which is non tangential this is what we discussed in the beginning of this lecture. For definiteness we have used normal direction that is all the partial derivatives or directional derivatives in two coordinate directions for a differentiable function knowledge of any two directional derivatives of course directions must be linearly independent that is enough to determine any other directional derivative as this map is a linear functional on r2 v going to dv u that is a directional derivative of u in the direction of v at the point p so p is fixed then the mapping v going to dv u at p is a linear functional which is fully determined once its values on a basis is known at any point on gamma 2 the tangential and normal directions are always linearly independent. Now let us look at computation of second order derivatives here we need to use a PDE we have determined the first order derivatives at all points of gamma 2 that is ux of fs gs and ui of fs gs we call them ps and qs on differentiating with respect to s we get p prime of s equal to uxx into f prime s plus ux y into g prime s similarly we get this expression for q prime s note here that the LHS p prime s and q prime s is known because p and s are known functions this is known f prime and g prime are anyway known so what are unknowns here uxx, uxy and uy y note that we are not making any distinction between uxi, uxy and uyx y because we are planned to compute all the derivatives and then propose a formal power series expansions for the solution therefore we are assuming that solution is smooth and for smooth functions the mixed partial derivatives do not depend on the order in which you take the derivatives so thus in conclusion 3 and 4 represent 2 equations 2 linear equations in the 3 unknowns therefore it will be nice to have one more equation so that we can hope to determine the unknown quantities namely the second order partial derivatives of u along the curve gamma 2. So the equations 3 and 4 feature 3 unknowns which are uxx, uxy, uy y it would be nice to have another equation satisfied by these unknowns so that we can hope to determine them the PDE the second order quasi linear PDE gives us a third equation because PDE is an expression for some combination of second order partial derivatives thus we have a uxx plus 2b uxy plus cuyy equal to minus d I have written in this form I have taken d to the other side because I wanted to write finally a system of linear equations for the unknown quantities where zeta s is fs gs hs ps qs. So the equations 3, 4, 5 may be written as the linear system in this linear system notice this is these are known quantities the matrix on the right hand side is known function of s therefore we can determine uniquely these quantities provided this determinant is non-zero. What is the determinant of this matrix let us denote it by delta of s because it keeps coming throughout this lecture it has this expression once you expand this determinant it turns out it is this is it is this. Observe that the system of linear equations on the last slide determine all the second order partial derivatives if delta s is not equal to 0 at points of gamma 2 wherever it is non-zero you can determine the second order derivatives at that point. So from now onwards assume that the above condition is satisfied. Now let us look go to the computation of third and higher order derivatives how the second order partial derivatives are computed if you look at pde determines a combination of second order derivatives of u along gamma 2 fine second thing is knowledge of first order derivatives of u on gamma 2 yields another two relations when we differentiate that with respect to s. So that is how we got the three equations and we could get all the three second order partial derivatives. Now if you want to repeat the above process what you need is a pde which gives a combination of third order derivatives how will you get that differentiate the pde if you differentiate the pde with respect to x or y you will get a new pde which has the third order derivatives in it fine and now second order derivatives we know on gamma 2 therefore if you differentiate that that will give you some more relations which involve third order derivatives of u. pde satisfied by third order partial derivatives we want to find so let us differentiate the given quasi linear equation with respect to x. So by product rule it turns out to be this notice the third order partial derivatives are appearing here and their coefficients are a to b and c which are exactly the coefficients in the given equation and this is nothing special for differentiation with respect to just x it will also be the same when you differentiate this with respect to y there will be a third order derivatives a different third order derivatives but coefficients are a to b and c even if you differentiate it 10 times even then you will get if suppose you differentiate 10 times then you get an equation which is a 12th order derivatives but with same coefficients a to b and c this will not change. So after differentiating we get this equation now we need to explain slightly what this notation stands for notice here a or a b c d they are all functions of x y u of x y ux of x y u of x y so that we are differentiating with respect to x that is why we have written this kind of notation. So let us introduce what this notation is so it is dou phi by dou x for any function phi a or b or c or d what it stands for is this dou by dou x of phi of x y u of x y ux of x y u y of x y so from the PDE that we obtained after differentiating the given PDE with respect to x it follows that a u triple x plus 2 b u double x y plus c u x double y namely this this is a known function along gamma 2 or this is a known function of s provided the rest of the things are known functions of s notice here a b c d are known functions and the second order partial derivatives u x x u x y u y y have already been determined along gamma 2 therefore they are known functions and on the next slide we are going to show that dou by dou x phi for any of these functions a b c d is a known function of s and then it follows that these are all known functions of s and hence this quantity is a known function of s that means this particular combination of u triple x u double x y and u x double y is known function of s. Take any function phi in a b c d what is this this by chain rule is exactly this differentiate phi with respect to x at this point zeta s differentiate phi with respect to the third variable which we are calling z and then that is u so u derivative of u with respect to x that is why ux and phi this is p this is q so phi p and phi q and this is ux therefore it is u double x this is u y therefore it is u x y therefore this is a known function of s the conclusion is that u a u triple x plus 2 b u x x y plus c u x y y is a known function of s. What is remaining is u triple y that is not appearing here for that we need to work separately we will discuss that later. The following system of equations holds for the third order derivatives at the point fs gs in gamma 2 see we knew u x x u x y u y y as a function of s these are been already determined therefore we can differentiate them d by d s d by d s d by d s. So, earlier for the first order derivatives we called p q as u x and u y now we can call r s and t but it will introduce new notations I want to avoid that that is why I am retaining it as it is but by chain rule this quantity is given in terms of a combination of third order partial derivatives of u so this we can get this expression. So, here all third order partial derivatives are evaluated at this point zeta of s the functions on the LHS are known because we know all the second order partial derivatives of u along gamma 2 so they are all known functions of s and hence their derivatives and on the right hand side we know f prime g prime the only thing we do not know is the third order partial derivatives which we are trying to determine. So, the system of linear equations given by 7a, 7b and 6 6 is the equation that we obtained after differentiating the given equation with respect to x and 7a and 7b are the first two equations on this slide and the right hand sides are all known functions I am not writing explicitly what they are because that is not important for us what we need to know is these are known functions these are the unknown functions which we are trying to determine and this matrix is interesting because exactly the same matrix that appeared in the computation of second order partial derivatives and we have assumed that is invertible therefore we can determine all these three derivatives. Now, we had an option of writing equations u triple y also featuring so we can write four equations but for this reason I avoided that this is convenient for us u triple y we will do similarly how to find u triple y okay so thus u triple x u double x y u x y y are determined along gamma 2 what remains is to find u triple y how do you get that many ways one of them is do the same procedure instead of differentiating the PDE with respect to x differentiate with respect to y and consider the last two equations on that slide where we had the three equations the equations which came out of differentiating second order derivatives along gamma so exactly same computations you repeat otherwise differentiate the given PDE with respect to y and then in that only u triple y will be unknown rest of the third order derivatives have already been determined so therefore you can determine u triple y of course you would need that the coefficient multiplying u triple y is non-zero what that can be done so the procedure described above can be continued indefinitely and all higher order derivatives of u may be determined there is no need to impose any more assumptions on gamma 2 other than requiring delta s not equal to 0 that is important of course you need all these functions to be infinitely differentiable now what are the curves gamma 2 for which delta s is identically equal to 0 this is a natural important question because such curves will prevent you from doing these computations okay we may not be able to determine the second order derivatives if delta s is identically equal to 0 or even if you are able to determine it is not unique so we do not say it is determined uniquely right so there is a trouble if delta s is identically equal to 0 what does that mean it just means this this is a equation equal to 0 for every s in i in example 2 PDE is uxx equal to 0 gamma 2 is x axis delta s is identically equal to 0 holds okay and we saw the trouble there either there is no solution or infinitely many solutions now how to find these curves how to determine these curves of course here the equation that we have is in terms of the parameter yes so now how do I find such curves in x-ray plane unfortunately it involves h s p s and q s okay because we have considered the quasi linear equations so therefore if you consider it is a linear equation it much easy okay assume that gamma 2 is a graph of a function for example x equal to xi of y then gamma 2 will be x equal to f s which is now xi s and y equal to g s which is s then this equation becomes this equation of course still zeta s is there so when the equation is linear then c zeta s does not depend on 5 quantities it only depends on the first 2 quantities which is x and y f s and g s so writing in terms of x and y we get this equation this is an ordinary differential equation first order but degree 2 there is a power 2 here so d xi by dy maybe one can compute using the formula of the solutions of quadratic equations and you are likely to get two equations likely to you may not get we will see that later in forthcoming lectures we will discuss about solutions of this equation. So let us summarize what we did a formal procedure to solve Cauchy problems breaks down if gamma 2 is a special curve one might ask why did we consider second order causality equation for this discussion simply because most general equations for which the method can be carried out are causality equations that is why we have done for causality equations for general non-linear equation we cannot do okay that a if you remember a depended only zeta s zeta s is already known the moment you compute the first order derivatives so whenever you differentiate as many number of times as you want the partial differential equation the highest order partial derivatives are always multiplied with a to b and c which are known functions and as a result the linear system that we may write from time to time will be the same whose determinant will always be delta s so that is the advantage since we could do we have done it for causality equations and the questions on special curves and their consequences will be important for the second order causality equations also we will come across them later interestingly we will run into special curves for PDE in a different context also which we will see in the next lecture in forthcoming lectures for linear PDEs we will try to find answers we will try to find answers to the following questions what are they do special curves exist for any second order linear equation if yes how to find them how many of them exist etc so in the next lecture we will take up another context where delta s makes an appearance delta s identically equal to 0 will become important thank you