 In this lecture we are going to consider second order partial differential equations linear which are identified as elliptic equations and try to obtain a canonical form for the same. Recall we have seen this example or one example of an elliptic equation namely the Laplace equation. So the canonical form for an elliptic equation we would like to resemble this. If you observe here in this equation u x y does not appear and u x x and u y appear with coefficients 1. So this is going to be the model for us for an equation of elliptic type. So let us consider the second order linear equation 12 which is elliptic, assume to be elliptic in a region omega, assume further that the coefficient a, b, c are real analytic functions. This is a requirement for the proof of our theorem. In the examples that you want to do you may simply follow the procedure, you may you will still be successful. So let x naught y naught belongs to omega. Now there is an open set containing the point x 0 by 0 and a change of coordinates such that the 2L gets transformed into this equation observe this part where the second order partial derivatives appear, w xi xi plus w eta eta looks like u x x plus u y y and no mixed partial derivative w xi eta does not appear in this equation. So this is what is called a canonical form for elliptic equations. So equation is given to be elliptic, it means b square minus a c is negative less than 0 on omega. Since the equation is elliptic we cannot have a to be 0, a of x 0 y 0 if it is 0 what we have is b square, b square strictly less than 0 is not correct because b square is always greater than or equal to 0, square of a real number is always greater than or equal to 0. Therefore necessarily a should be non-zero and of course c also cannot be 0 for the same reason. So neither a nor c can be 0 at the point x naught at any point in omega in particular at x naught y naught and we are assuming they are real analytic definitely continuous. So by continuity of the function a, the function a will be not 0 in some open set u which contains x 0 y 0 by continuity. Recall the change of variables xi eta are given by two functions phi and psi if there is if this gives rise to change of variables you can invert back x and y you can write in terms of xi eta we use capital phi and capital psi for that and a function of x y can be identified with a function of xi eta and they satisfy these two relations and the 2L equation second order linear equation gets transformed to this equation where the coefficients a, b, c alone are listed here because they are the only things which are important as far as the type of an equation is concerned. So what do we need for proving the theorem? We want the w xi xi and w eta eta should appear with coefficients 1 in particular a must be equal to c and equal to 1 and b is 0 once b is 0 and a equal to c we can divide with a then I will get w xi xi plus w eta eta anyway. So the condition is a equal to c and b equal to 0 that is enough. So a equal to c this is a and this is c. So a equal to c means this equation must be satisfied and b equal to 0 means this equation is satisfied. The system of equations for finding phi and psi a remark on that recall that in the hyperbolic case when we want to find phi and psi the equations were decoupled right they were decoupled phi and psi could be solved separately. In the parabolic case they are weakly coupled the equations for phi and psi are weakly coupled in the sense that the equation for phi in did not involve psi at all. Unfortunately once we solve for phi and when we go and substitute in the second equation the equation reduces to the identity 0 equal to 0. Maybe it is a good thing because now we can choose psi arbitrarily as long as a Jacobian of phi and psi is non-zero. This can also be explained in following way. Once you solve for phi which is solution of the equation a of psi, eta equal to 0 we need not even solve for b psi eta equal to 0 because it is automatically satisfied by the invariance of the classification type under change of coordinates. Not only that it is satisfied by any function psi b of psi eta equal to 0 is satisfied by any function psi. Therefore we had lots of freedom in choosing psi the only requirement was that the Jacobian of phi and psi is non-zero. But for elliptic equation what is happening in the here also you have phi and psi mixed here also you have phi and psi mixed therefore it is a strongly coupled system for phi and psi of first order non-linear PDEs. We can overcome this difficulty by using the assumption of the real analytic city of ABC and some complex variable techniques. You will see that a crucial part of the proof we will skip we will not do the proof. This system may be rewritten as this right the first equation I simply transfers all these terms to the left hand side and I take a common so that I get phi x square psi x square to be common so that I get phi x phi y minus psi x psi y and C common so that phi y square minus psi y square it is exactly the same equation rewritten. Second equation I am just multiplying with I with because I have some idea what I want to do in the next slide that is why I put I otherwise the same equation I multiplied with I it tells us that some complex things are going to enter. So define a complex valued function capital phi by phi of x y plus I psi of x phi. So this system the system that we want to solve for phi and psi is equivalent to this one single equation now a phi x square plus 2 b phi x phi y plus C phi y square equal to 0. This equation we have seen earlier this is the same equation that we solved while determining canonical form for hyperbolic equations of course the difference was there that we could factorize as real equations factors were real but here it will not happen it leads to PDGs with complex coefficients given by this equation and this equation. In the case of hyperbolic equation there was no I first of all and inside thing was b square minus ac. So these are the 2 equations we have now a fact the system of equations has solutions near x 0 y 0 this is where a b c are real analytic functions is used this we can observe if phi is a solution of the first equation remember a b c they are all real valued functions therefore if phi is the solution of the first equation phi bar the conjugate of phi that let us call it is capital psi that is a solution to the second equation therefore it is enough to solve one equation essentially there is only one equation right but that will give us what we want because the phi is proposed as small phi plus I times psi. So you can identify a real part and imaginary part and hopefully that forms a coordinate system that defines a coordinate system therefore phi and psi are constant on the 2 complex characteristics these are right hand side is a complex valued function okay we have not studied in our Picard's theorem how to solve such equations okay this is where it is important that we use our hypothesis and show that solutions exist define small phi of x y equal to real part of capital phi small because we have obtained phi after this right this phi exists is what somebody told us this fact so I have phi and then real part I will call a small phi and imaginary part I call psi and this defines a coordinate transformation near the point that is left as an exercise very easy and the transformer equation has a required form because we made what we want right A equal to C and B equal to 0. So therefore we will get this if you want to see a detailed proof please look at this book by PR Garabidian on partial differential equations you will find full details let us solve an example here you see we do not care whether it is analytic or not of course here these functions here it is a constant function 1 here it is 1 plus y square whole square it is a polynomial here also it is a polynomial of course it does not matter we are worried only about coefficients of u x x u y y and u x y. So B square minus AC is negative strictly less than 0 at every point therefore the equation is of elliptic type everywhere in the plane or 2. So let us transform now the given equation into it is a canonical form. So we need to solve these ODE's dy by dx equal to plus or minus i into 1 plus y square. So we need to solve this ODE i times this because minus i times does not matter it will be the conjugate. So this solution is given by tan inverse y minus ix equal to constant okay please accept this that this is a solution okay and then real part is x imaginary part tan inverse y or y square star here of course if this is constant I can multiply this with i also again right. So therefore it does not matter what I call the variables phi and psi in the theorem that we presented this was called phi and this was called psi but we are interchanging here it does not matter because there is no preference for the variable x or y or psi or eta. So therefore you propose u of x y equal to this w of x comma tan inverse y differentiate ux is x appears only here so it is w psi and derivative of x is 1. Uy will be w eta and derivative of this is 1 by 1 plus y square. So you can continue like that compute the derivatives go back and substitute in the given equation we get this. So that is a canonical form of the given PDE. So summarizing what we did is that we have presented a method to reduce a second order linear PDE which is of elliptic type to its canonical form we have of course assumed very high assumptions on the coefficients ABC but you may generally if you are given a partial differential equation of elliptic type you want to find its canonical form you may simply follow the procedure do not bother about the hypothesis checking for the theorem and you will still be successful if you are able to solve the ODEs which are coming on the way and we have seen the method is successfully implemented in the example. Thank you.