 In the last lecture we have discussed methods to find general solutions to linear and semi-linear equations. In this lecture we take the method one more step forward namely to deal with quasi-linear equations. We are going to describe Lagrange's method for finding general solutions to quasi- linear equations. We start with a brief review of the last lecture, lecture 2.3 where general solutions to linear and semi-linear equations have been obtained. Basically a method was described. Now we are going to describe in today's class Lagrange's method for finding general solutions to quasi-linear equations. How the ideas from linear and semi-linear can be extended to the case of quasi-linear equations whether it is possible or not we are going to see and which part of this method for linear and semi-linear can be extended which cannot be we will see that as well. So let us recall just to reinforce the notations L stands for the linear equation axyux plus bxyuy equal to cxyu plus dxy. S L stands for semi-linear equation where the left hand side is same as that of the linear equation that means the manner in which the first order partial derivatives appear is exactly the same. However the right hand side can depend non-linearly on u c of xyu. Now q L stands for quasi-linear equations. Here the coefficients of ux and uy can depend on u also. So a of xyu ux plus b of xyuy equal to c of xyu and the hypothesis that we will be working under is abc or c1 functions on omega 3 omega 3 the subset of R3 open subset of R3 and as usual we require that both a and b do not vanish simultaneously at each of the points of omega 3. So this is quasi-linear equation. As you observe the first order partial derivatives the coefficients in both L and S L they depend only on x and y whereas in quasi-linear equations they depend on u as well. This is the difference between semi-linear linear and quasi-linear equations. So let us review lecture 2.3 now where we have obtained general solutions to linear and semi-linear equations. Basically a method was described now it is up to us whether we are able to implement it or not that is silent upon but it is possible that is what the method says. What are the important steps in obtaining general solutions? There is a change of coordinates involved xy to xi eta because in L and S L both ux and uy appear. So we thought that we will eliminate one of those partial derivatives but you know ux and uy are there x and y coordinates are there coefficients a, b are given we cannot suddenly make it 0. Then we thought let us change coordinates and then find out if there is a possibility of finding change of coordinates where after the transforming the PDE the new PDE in the new coordinates will feature only one derivative maybe with respect to xi or eta only one of them. That was the idea therefore there is a change of coordinates in the background. Remember change of coordinates works in xy plane that is in omega 2 some subset of omega 2 omega 2 there we have both xy and xi eta coordinates. And then we chose psi to be a solution of this linear homogeneous PDE and then we chose a phi such that this Jacobian is never 0 then the LHS of L and SL which is this now becomes this it features only the derivative with respect to xi and the PDEs will then become ODEs which are then solved that was the idea. Now after changing the variables back to xy from xi eta we get the solutions as functions of x and y. Now how to extend this strategy to quasi-linear equations? A, B are functions of xy only they played an important role in obtaining change of coordinates. Therefore the new difficulty in implementing this strategy is that for quasi-linear equations A, B are also functions of the variable z. So this results in non-linear PDEs for psi equal equivalently phi because we have as we observed the equations are the same and only one of them is useful so we can only find one function because if you choose both phi and psi as solutions of the same equation then we will get that Jacobian will be 0. So that is the reason we chose psi first and then we choose phi as anybody such that the Jacobian is non-zero. So now it is clear the new difficulties A, B depend on x, y, z unlike L and S L this we are mentioning for the completeness sake in the sense that phi psi satisfy non-linear PDE but it is useless because change of coordinates now have to be done in x, y, z coordinates not in x, y but the PDEs only in x, y variables. So therefore we do not expect change of coordinates can be done. Therefore in what other way we can extend the strategy? Let us discuss that now. In L and S L psi was chosen such that psi equal to k represented a one parameter family of solutions to t by dx equal to b by a. This is what we did. Now in the quasi-linear case what happens is B of x, y, z, A of x, y, z will come. So necessarily we have to bring in some equation for z we will come to that. So psi of x, y equal to k represent base characteristic curves. This is the equation for base characteristic curves after eliminating the parameter t. Therefore for quasi-linear equations we cannot determine base characteristics directly that we know because of the presence of z in both A and B. So for quasi-linear equations looking at characteristic curves is thus natural. So this is the way we are going to extend the ideas. We are saying that for linear and semi-linear looking at base characteristic curves helped us in obtaining a general solution to L and S L. For Q L we have to look at full characteristics. So non-parametric form of the equations for characteristic curves this is after eliminating the parameter t is d by dx equal to b by a, dz by dx equal to c by a. This is the idea of Lagrange's method but no change of variables as expected. So let us now describe Lagrange's method for quasi-linear equations. So Lagrange's method the main idea is this before we present that we will assume that A is never 0 on omega 3. So Lagrange's idea is to consider the two parameter family of characteristic curves corresponding to quasi-linear equation given by dy by dx equal to b by a and dz by dx equal to c by a. These are the equations of the characteristic curves. Assume that the two parameter family of characteristic curves are given by intersection of two families of surfaces phi of x, y, z equal to c 1 and psi of x, y, z equal to c 2. This is the assumption. We will see in examples that we can get this. So finding the really the phi and psi is the most challenging part of this Lagrange's method. Okay let us have let us state the theorem of the Lagrange's method. Assume that phi equal to constant and psi equal to constant represent characteristic curves for ql and grad phi and grad psi are never parallel to each other. At all those x, y, z which satisfy both these equations at those points grad phi is never parallel to grad psi. Let f be an arbitrary c 1 function defined on R2 such that f psi square plus f eta square is not equal to 0. In other words the gradient of phi, sorry gradient of f is never 0 at every point in R2 because domain of f okay. If you do not have R2 here and some domain of f then we require this for every psi eta belonging to domain of f psi, f eta is not equal to 0, 0. They do not simultaneously vanish. Assume that f of phi of x, y, u, psi of x, y, u equal to 0 this equation defines a c 1 function x, y mapping to u of x, y. That means this is an implicit equation for u and that can be solved. u can be solved in terms of the other 2 variables. For x, y belonging to some open subset of omega 2. As usual we do not demand that everything should happen on omega 2 because typically these the existential results which deal with such assertions is implicit function theorem. That is once again a local theorem. So for x, y belonging to an open set we demand that f of phi of x, y, u, psi of x, y equal to 0 gives rise to a function u of x, y okay. If f, phi, psi or c 1 automatically u will become c 1 okay. That is we are discussing now whether you can always assert this that is implicit function theorem. But as far as this theorem is concerned we assume that such a thing exists. Then u is a general solution of the quasi linear PDE QL. First observation is ABC is parallel to grad phi cross grad psi. Observation 2 is such a PDE is satisfied okay. What is grad phi grad phi cross product with grad psi is this okay. In other words this is this right. This is the first component. This is the second component and this RHS is the third component okay. If this is parallel to ABC what does that mean? This vector is some constant times ABC therefore that let us say alpha ABC. So you can substitute this alpha a, this alpha b, this alpha c. Cancel alpha what you get is AUX plus BUI equal to c. So therefore these two observations prove the theorem. So therefore the proof of the theorem follows from the following two observations. So what remains to prove is these two observations. Observation 1 what is observation 1? ABC is parallel to grad phi cross grad psi. So phi equal to c 1 and psi equal to c 2 describe characteristics curve for QL. This is the assumption. What does that mean? It means that given any characteristic curve x t, y t, z t parameterized by t belonging to some interval j of R, x t, y t, z t satisfies both the equations. What are the equations? Phi of x t, y t, z t equal to c 1 and psi of x t, y t, z t equal to c 2. It lies on the surfaces okay. That is the meaning of saying this and this together describe characteristic curve. It means if your characteristic curves they lie on the intersection of these two surfaces. So it means that there is a c 1 and c 2 such that phi of x t, y t, z t equal to c 1 and psi of x t, y t, z t equal to c 2. Now once we have this we differentiate with respect to t both these equations and we will get ABC dot grad phi is 0 and ABC dot grad psi is 0. So this is what we have but grad phi cross grad c is not equal to 0. That is what we have assumed. Therefore ABC must be parallel to grad phi cross grad psi. If a vector is orthogonal to two vectors let us say u dot v is 0, u dot w is 0 then u is parallel to v cross w if v cross w is non-zero. So we have used that and we concluded ABC is parallel to grad phi cross grad psi. This holds at every point x t, y t, z t on the characteristic curve. So this is the notations we have used. This stands for this particular determinant. This is the Jacobian of phi psi with respect to the variables y, z and so on. Observation 2, what is observation 2? Observation 2 is such an equation is satisfied. This equation is satisfied. Let phi and psi be c 1 functions. This is how we write phi is defined by phi of x, y, z just to denote the variables for the function phi. Later on we are going to substitute in place of z u of x, y. But this is the function of 3 variables. So it is defined on an open subset of R3. It is very much useful when we apply chain rule so that we do not have confusion. So f from R to R be an arbitrary c 1 function and we denote this by f is f of xi, eta. Assume that this equation defines a c 1 function. That was part of the assumption. Now differentiate this equation with respect to x and y. So if you want to differentiate this equation with respect to x, x is present in both of them. Therefore, differentiate f with respect to xi at this point phi, psi which are the arguments I have not written. Then differentiate this with respect to x. But here x appears here in the first coordinate and also in the third coordinate. So differentiate phi with respect to x plus x with respect to x is 1. That is why it is phi x plus phi with respect to the third one. The third variable we have denoted by z. So it will be phi z into u is there. Differentiate u with respect to x. So you get u x. Similarly, you differentiate here also. You get f eta because second variable of f was called eta. So f eta at this point into derivative of psi with respect to x plus derivative of psi with respect to the z variable into derivative of u with respect to x. That will give you the first equation here. Similarly, we get the second equation if you differentiate this with respect to y. So here we should have written u of xy and here u of xy. That is the meaning here. It defines a c 1 function. That means I can substitute here u of xy, u of xy. And that equation would be valid on a subset D of r2. So the system on the last slide, it is a linear system for f xi and f eta. f xi and f eta. So you write f xi, f eta as one column. Then this is the first row a 1 1, first 1 1 entry, 1 2 entry, 2 1 entry, 2 2 entry. This is what we have. Now we have assumed that both f xi and f eta cannot be 0 simultaneously. Which means this system has a non-trivial solution, a non-zero solution. This homogeneous system, it has a non-zero solution. It means that the determinant of this is 0. And if you expand, this is what we get. So one of the hypotheses of the theorem is grad phi is never parallel to grad psi. What does that mean? What is the geometric interpretation? For each c 1, c 2, the surfaces phi equal to c 1 and psi equal to c 2 intersect transversely. And their intersection cannot be another surface. Because normals do not have the same direction. Normals have different directions. Therefore the tangent planes will have different directions. At every point which is common to both the surfaces phi equal to c 1 and psi equal to c 2 at such a point, look at the tangent plane for phi equal to c 1 and tangent plane for psi equal to c 2, they do not coincide at all. That is the assumption we are making. The grad phi is never parallel to grad c anywhere. So Lagrange's method provides a general solution implicitly by this relation. Because we assume that you can be solved for in terms of x, y. In such a case, we define that to be a general solution. Where f is arbitrary function, we have just make sure that phi and psi take values in the domain of f. If you take an f in R2, then we do not have to take even that much care. Once phi and psi have been written mind, this is where the catch lies. This itself is the drawback of the method, finding phi and psi. Let us look at an example. Using Lagrange's method find general solutions to solution to this PDE, x u x plus y u y equal to u. Actually this is a linear equation. Characteristic ODEs are given by, this is another way of writing dx by dt equal to x, dy by dt equal to y, dz by dt equal to z. After since t is not present, this is another way of writing dy by dx equal to y by x, dz by dx equal to z by x. So you integrate dx by x equal to dy by y, you get log mod x equal to log mod y plus constant for some constant. And that gives us that mod x by y is e power k. And that implies x by y is c 1, where c 1 is non-zero. Because e power k will never be 0, no matter what k is e power k is never 0. So x by y equal to c 1. Similarly on integrating the other equation dy by y equal to dz by z, we could choose any combination. We just want two functions phi and psi. So this is one solution, x by y equal to constant. It is a solution of this particular equation dx by x equal to dy by y. Now we will look at dy by y equal to dz by z. We get similarly y by z equal to constant. So now what we want is phi and psi such that phi equal to c 1, psi equal to c 2 together represent characteristic curves. So now by Lagrange method says f of phi, psi equal to 0. So therefore, this is represents a general solution. Let us look at another example. Unfortunately, this is also linear. We will discuss some quasi-linear equation in the tutorials. So characteristic ODEs are given by dx by y minus y equal to dy by x dx by a that is minus y dy by b that is x dz by 0. Now on integrating this equation, we get x square plus y square equal to constant c 1 positive. And integrating this we get z equal to constant. Therefore, Lagrange method says general solution is given by f of x square plus y square comma z equal to 0, f is arbitrary. If I take a specific choice of f which is eta minus g psi, what I get is u equal to g of x square plus y square. In other words, this solution if you observe, this is constant on each circle with center at origin, x square plus y square equal to constant, positive constant will give you u is constant on all circles with center at origin. Please note that if you have integrated this differently from what is done here, you will get f of some other two functions equal to 0. And those two functions, these two functions will be related somehow. They will be functionally dependent. I do not want to use these terms and discuss this further. But as long as you get phi equal to constant and psi equal to constant representing characteristic curves, capital F of phi comma psi equal to 0 will always represent general solution provided we can solve for z in terms of x and y. For example, if I chose this f, of course, this is solvable u equal to g of x square plus y square where g is an arbitrary function. So let us summarize what we did in this lecture. In lecture 2.3, linear and semi-linear equations were transformed to ODEs by a change of variables. General solutions were obtained. In this lecture Lagrange's method to solve QL was presented. How ideas from simpler cases of L and SL extend to QL were looked into. And we have presented two examples to concretely illustrate the general algorithm, which is given for finding general solutions to QL. And we cannot do this kind of general solutions to non-linear equations. So beyond QL, our theory does not go. So to solve even such equations, we will try to solve using what is called method of characteristics. In the next few lectures, we will be presenting the method of characteristics and its preliminaries initially. And then we use that method to solve quasi-linear equations first and then once again extend those ideas to solve general first order non-linear equations. Thank you.