 In this lecture, we are going to see given an equation of parabolic type, how to get it into its canonical form. For parabolic type equation, remember we want to model it after heat equation. How does heat equation look like? It is ut equal to uxx, right. So in the second order part, for example, let us write down ut equal to uxx, this is the heat equation. In this, if you look at the part where the second order partial derivatives appear that is simply uxx. In particular, uxt is not present and utt is also not present. So this is what we are planning to do. Given an equation of parabolic type, equation which is parabolic at every point in some domain, we want to change coordinates so that the transformed equation will feature exactly one such second order partial derivative where it is with respect to the same variable x and x. The other variable is completely absent. So let us state the hypothesis which is required to do this. There is not really any hypothesis here. It is very standard, very simple and what is to be expected. So let this equation be parabolic in a region omega of the xy plane. This is a second order linear equation. We assume these coefficients ABC are at least continuous. We need that and take a point in omega. So the theorem as before for the case of hyperbolic equations, it is going to be of local nature. That is why we take a point x0, y0 in the domain omega where the equation is parabolic. Conclusion is that there is an open set containing this point x0, y0 and a change of coordinates x, y going to xi, eta that is a new coordinate system. When the equation 2L is expressed in the xi, eta coordinates it will look like this. If you notice here w, eta, eta these are only second order partial derivative which appears. So none of the second order derivatives involving the variable xi appears. So given that the equation is parabolic in omega, what does that mean? b square – ac is 0 in omega. So to have a second order partial differential equation, we need that at least one of the functions ABC is not 0 at every point of omega. So we know this. All of them cannot be simultaneously 0. We are not allowing such coefficients. Therefore either a is not 0 or c is not 0. If it does not happen what will happen? a and c both of them are 0 at x0, y0. The moment a, c is 0, this equation tells b square is also 0 at that point. That means b of x0, y0 is 0. It means all the 3 namely a of x0, y0, b of x0, y0 and c of x0, y0 are 0 and that is not allowed. Therefore we have this. Either this is not 0 or this is non-zero. At least one of them happens. So assume without loss of generality that a is non-zero and here is a twist and c is also not 0. We can arrange this. What we are concluding in the previous slide is that given any equation of parabolic type at any point in the domain x0, y0 at least one of the a or c is non-zero. But now we are saying that we can make it happen that both of them are non-zero but after a change of variables. Imagine a is non-zero and c is 0 at the point x0, y0. Then let us introduce a change of coordinates x, y going to x tilde, y tilde where x tilde is defined as a function of xy is just x and y tilde as a function of xy is x plus y. So then the equation 2L takes the form where the coefficients of wx tilde, x tilde and wy tilde, y tilde are non-zero at this point x tilde equal to x0, y tilde equal to x0 plus y0. We have already introduced change of coordinates and how a PDE changes under change of coordinates into a new equation. Use those formulae and conclude this very simple exercise. I will not be giving the details of this. So therefore we can assume that a is not 0 and c is not 0. So there is one more case it may happen that a is 0 and c is not 0 then also we can introduce this change of coordinates such that we have wx tilde, x tilde and wy tilde, y tilde their coefficients are non-zero that can be arranged. Why are we doing this because in our proof we need that both a is non-zero and b non-zero because if a is non-zero it may happen that b is 0 right what all the condition for parabolicity we have is b square minus ac is 0. So a is non-zero c may be 0 in which case b is 0 we do not want that in our proof our proof will not work we require that both a and b are non-zero in a neighborhood of x0, y0. So the moment we are now we have decided we can there is no loss of generality in assuming this. After assuming this a is a continuous function and it is not 0 at a point therefore in an open set it will continue to be non-zero because of the continuity same thing will hold for c and hence for b also. If c is also not 0 what do we have b square minus ac is 0 a is not 0 and c is not 0 what we know is this b square at x0, y0 b square minus ac is 0 therefore I can write this is equal to ac this is not 0 that means b is also not 0. Once b is also not 0 at the point x0, y0 in an open set it will be non-zero there are after all three functions a, b, c so we can take the same open set on which all of them are non-zero and recall the change of variables psi in data by the functions phi xy and psi xy then this is the inverse functions for that and the function u xy in terms of w is w of phi xy psi xy and w in terms of u is given by this equation. And we have seen that the 2L the second order linear equation under this change of coordinates becomes this where the coefficients a, b, c which matter for the type of a equation are mentioned here. So for proving the theorem what do we need? We need that only this should appear so c should be non-zero a and b must be 0. Suppose we choose the coordinate system such that a and b are 0 once a and b are 0 c cannot be 0 because if all of them are 0 there is no differential equation we are not allowing that. So a, b, 0 automatically means c is not 0 once c is not 0 you can divide this equation with c and get w eta eta with coefficient 1. So this is what we want a and b to be 0 and once that happens c is automatically non-zero because at least one of the three functions a, b, c are non-zero at every point. So thus we need to find the functions phi and psi such that phi satisfies this equation this is basically a equal to 0 and phi and psi satisfies this equation that is the equation b equal to 0. If you have observed this equation the first equation involves only phi and second equation involves both of them. So we may say that there is a coupling between phi and psi the system couples both of them but in some kind of weak coupling because one equation does not have psi at all the other one has psi but then once you know phi it is an equation only for psi this is something similar not exactly same analogous to what we have in linear algebra. We have a linear system of equations which are triangular okay they are coupled but nicely coupled it is like that. So since the first equation involves only phi we may solve for phi as we have done in the previous lecture exactly same way we factorize but now what happens b square minus ac is 0. If once b square minus ac is 0 this factor in this bracket parenthesis is same as this one. So this equation reduces to this. So therefore we need to solve phi such that a phi x plus b phi y is 0 and choose psi arbitrarily okay we do not want to go back and solve that equation we will see in a moment why that is so. What all we want is a certain change of coordinate system that is all we want and if you notice if a is 0 b is automatically 0. So there is no need to check that that is satisfied b square minus ac is 0 right. So if a is 0 then ac is 0 therefore b square is 0 therefore b must be 0. So there is no need to check the equation for b. There is no need to check the equation for b because b of xi eta equal to 0 is satisfied by any function psi as long as phi satisfies the equation a of xi eta equal to 0. That is why we have lots of choices for psi because the only constraint on psi now is that the Jacobian of phi and psi is non-zero which will give that xi equal to phi xy eta equal to psi xy will define a coordinate transformation. So we choose psi arbitrarily with only one constraint namely this should give rise to a change of coordinates which means certain Jacobian non-zero and with the choice of phi as above we have a equal to 0 right and consequently b is 0 because the equation type is invariant under change of coordinates. Let us look at the solutions of this equation a phi x plus b phi y equal to 0 system of characteristic ODE is given by dx by dt equal to a dy by dt equal to b dz by dt equal to 0. So this says that any solution of this PDE is constant along solutions of the base characteristic solutions of this ODE along each base characteristic solution is going to be a constant. So assume that phi of xy equal to k represents a one parameter family of solutions to this ODE. Of course this ODE represents base characteristic curves. On differentiating this equation phi of xy x equal to k we get a relation this is by chain rule we get this identity and then we get dy by dx equal to b by a and on other hand we get minus phi x by phi y. Now we need to choose psi such that this transformation xy going to phi xy psi xy is a non-singular transformation namely the Jacobian is not equal to 0 at the point x0 y0 we are looking at only x0 y0. So applying the inverse function theorem we conclude that psi eta equal to phi xy psi xy defines a new coordinate system here x0 y0. We have to find psi such that this condition is satisfied non-singularity of this transformation that rest will follow. Now to achieve this phi and psi need to satisfy this Jacobian to be non-zero then you can apply inverse function theorem right if the Jacobian is non-zero at x0 y0 then you can invert the transformation which is coming here xy going to phi xy psi xy can be inverted back so psi eta going to capital phi of psi eta and capital eta of psi eta we can do that sorry capital psi of psi eta. So what do you mean by this is equal to 0 or not equal to 0 it just means that this quantity phi x by phi y and psi x by psi y are not equal because this is 0 means they are equal not equal to 0 means they are not equal. So we have to find psi okay now we will use this condition this is not equal to phi x by phi y there are infinitely many such choices but we make a nice looking choice for theoretical purpose when we are dealing with an example we may not necessarily do like this. So this is what we know for phi x by phi y. So therefore we choose psi such that psi x by psi y is a by b okay if you look multiply these 2 things you get minus 1 minus b by a into a by b is minus 1. So this means that the curves corresponding to phi equal to constant and psi equal to constant are orthogonal families of curves. So with these choices of phi and psi inverse function theorem guarantees that psi eta defines a coordinate transformation near the point x0 y0 and therefore a and we are 0 C is definitely non-zero C cannot be 0 now you divide the equation with C and you get what you want W eta eta plus these terms do not matter what they are this alone matters and this is known as the canonical form for a parabolic equation. So canonical form for a parabolic equation means the second order derivatives exactly one of them appears and one variable is missing which means it should be W eta eta okay and W eta psi W psi psi do not appear. Let us look at an example here of course all the derivatives of you are appearing it is not clear so we have to find what is the type of this equation a is x square b is minus xy C is y square. So therefore bc minus a square is 0 so the equation is parabolic at every point in R2 parabolic everywhere in R2 this equation is not well defined at 0 0 xy equal to 0 0 because all a b c vanish therefore we do not want to include such a point in our consideration. So we will find its canonical form in the right half plane and this restriction is a technical one as I pointed out because the method followed here is valid only if we avoid x equal to 0 or y equal to 0. So canonical form can be found in any of the half planes not necessarily right half plane you can do in the left half plane or you can do upper half plane or lower half plane. So let us transform the equation into its canonical form in the right half plane. So we need to find new coordinate system for that we have to solve this will give us a phi dy by dx equal to minus y by x will give us phi of xy and then we have to make a choice for psi that is a procedure in the parabolic equations. So solutions are very easy here so they are given by xy equal to constant. So we need to choose psi right such that Jacobian is now 0 once you plug in the values of phi in this this should be phi actually not psi and eta but phi x phi y. So here it will be y here it will be x so y and x are there you need to fill with somebody so that Jacobian is now 0 I have chosen it to be x because I will get this and x is not 0 in my domain in the right half plane x is never 0. So we introduce this change of coordinates on differentiating this equation u xy equal to w of xy comma x and compute the derivatives and then go back and substitute in the given equation. So u x, u y compute like this u xx is this and u xy is this u yy is this please do the computations on your own pause the video do the computations and substituting these values in the given equation we will get this of course this equation still has x and y so it can be written in completely in terms of psi eta on expressing x and y as functions of psi eta that is nothing but writing the inverse function. So x equal to capital phi psi eta is eta y equal to capital psi psi eta is psi by eta. This transformation makes sense because eta is never 0 in our domain eta is 0 if and only if x is 0 and we are working in a domain where x is not 0. So therefore the transformed equation will be this and we want this to be 1 so divide with eta square so we get this expression. So this is the canonical form of the given PDE notice only w eta eta appears w eta xi or w xi xi do not appear that is how the canonical form is identified for parabolic equation. So what we did today is we have devised a method to reduce second order linear PDE which is a parabolic type to its canonical form and it was implemented successfully in an example. So in the next lecture we will take up equation of elliptic type and get the canonical form for such equations. Thank you.