 In this lecture we are going to find a canonical form for an equation which is of hyperbolic type. In the next two lectures we will be doing the same but for equations which are parabolic or elliptic types. So canonical form for hyperbolic equations what is it? We know what is hyperbolic equation we have defined. What is a canonical form? If somebody asks you give me an example of hyperbolic equation immediately you give an example of the wave equation. So therefore finding canonical form for hyperbolic equations consists of the following. You take the equation which is hyperbolic in a domain and then you want to do change of coordinates and then convert the PDE into the new coordinate system. But after doing this the part which features the second order derivatives looks like the wave equation, the one in the wave equation. We are going to do the canonical forms only for linear equations. For quasi linear equations it is very difficult so we will not do that. We state what we are going to see as a method as a theorem. So hypothesis consider the second order linear equation 2L which is given by this. Given that this equation is hyperbolic in an open set region means open and connected set in open set omega of the xy plane. So throughout the domain omega the equation is a hyperbolic type. Now take a point x0, y0 in omega that means take a point in omega. Conclusion is there is an open set containing the point x0, y0 and a change of coordinates defined on that open set such that the equation 2L is transformed into this equation. If you notice here and compare with the equation that we saw in the last lecture after change of variables we had a, b, c. So what is missing is a and c, a and c are 0 and b equal to half because 2b is what appeared so that is equal to half. So coefficient of w xi eta is 1, w xi xi and w eta do not appear in the equation. Of course it does not look like the wave equation that is understood ok, wave equation is not like u x, u x t equal to something, wave equation is u t t minus u x x, we will come back to that later. So given that the equation is hyperbolic that is b square minus a c is positive at every point of omega. We know that at least one of the three functions a, b, c is non-zero at every point. What happens if all of them are 0, pd is not defined right, it is not a second order pd. So we have to add this assumption whenever we consider even quasi-linear equation or linear equation we have to assume that none of the three functions a, b, c simultaneously vanish. That is an important assumption which we forgot to add now we add. In the proposed canonical form only mixed partial derivative appears we saw that. So if a and c are 0 at the point x0, y0 then definitely b is non-zero because if both a and c are 0, b square must be positive in particular b is non-zero. But it is of no use or help you may be thinking that a, c are 0. So in the equation a, c is not there what you have is only b of x0, y0 into u xy and it is b x0, y0 is non-zero divided by it yes that is fine you have a equation which looks like canonical form but remember it is only at the point x0, y0 you cannot assert this will happen in an open set containing x0, y0 because these conditions do not guarantee that if a function is 0 at some point you cannot say it will be 0 in a neighborhood of that. You can say of course for this if somebody is non-zero at a point if the function is continuous then definitely you can say it will be non-zero in a neighborhood. Yeah we have to place assumptions on a, b, c we have to assume that minimum we have to assume that they are continuous functions. You will see as and when we see there is a need to introduce hypothesis on a, b, c we will do that. Now we are saying assume without loss of generality this WLOG stands for without loss of generality that a of x0, y0 is not equal to 0 or c of x0, y0 is not equal to 0 because if they are 0 we can do something and get at least one of them to be non-zero. So if they are 0 indeed then introduce a change of coordinates xy going to x tilde, y tilde given by x tilde is given by as a function of xy as x plus y and y tilde is x minus y. Now if you look at the equation that will result by this change of variables the equation will have W x tilde, x tilde and W y tilde, y tilde they are non-zero. At the point x0 tilde which is given by x0 plus y0 and this point is x0 minus y0 because we have to talk about x tilde, y tilde right that coordinate system. So x tilde equal to x0, y0 that is a point y tilde equal to x0 minus y0 that is the point. So x tilde, y tilde what is the point precisely maybe this is what we have to write x tilde, this point is x0 plus y0 this is a known point because x0, y0 are known at this point. So therefore we may assume that at least one of the two quantities is non-zero, if they are 0 we will exactly first implement this change of variables we will get to an equation. In that equation we have these non-zero therefore we can as well start with assuming that there is no loss of generality in assuming that a is not 0 or c is not 0. So let us assume that a is not 0, a of x0, y0 is not 0 and we need to now assume that a is continuous so that there will be an open set in which a is not 0 containing that point. So assume that a is continuous, a, b, c are continuous functions. So recall change of variables they are going to look like this and the function u will become w in the new coordinate system and the connection between u and w is given by these equations. And the second order linear equation transforms to this and we have listed on a, b, c in the last lecture we said d, e, f, g does not matter but if needed I can always compute but our requirement now is that I want to make a and c as 0 and b as 1 by 2 or non-zero that is good enough. Once b is non-zero then I divide everything the entire equation with 2b so that I get 1 times w zeta. So for proving the theorem we need to find a coordinate system zeta so that a and c are 0 and we know that the type classification type of an equation does not change under change of coordinates. Once a and c is 0, b will be non-zero because b square minus ac is positive and ac is 0 therefore b square is positive that means b is non-zero. So we need to find phi and psi satisfying a phi x square plus 2b phi x phi y plus c phi y square equal to 0 and a psi x square plus 2b psi x psi y plus c psi y square equal to 0. This is a, this is c that is expression we get. What are these? These are first order PDEs, non-linear PDEs. Equations are identical if you see phi solves this or psi solves this does not matter it is a same equation. So phi and psi solve the same equation. So we need to solve for phi psi using only one equation. Now we write this equation like this if you multiply and divide with a you get back this so you can write like this and we have a license to do this that is why we have shown in the beginning that a is not equal to 0 in that neighborhood some neighborhood of x 0 by 0 therefore 1 by a is meaningful for this we have done that exercise. So a function phi is a solution to the above equation as long as it satisfies either this or this any one of them if it satisfies then the product will be 0 right if phi is such that this first part is 0 product is 0 therefore this is 0 and these equations are not similar I mean they are similar but not exactly same because there is a minus sign here there is a plus sign here that is what helps us in getting phi and sign. So whenever phi is a solution to any of these two equations it will be a solution to this equation because this is just a factorization of this it is the same equation expressed differently. So we choose phi to be a solution of the first equation from here and we choose psi to be a solution of the second equation. This choice means we have a change of coordinates why is that we have to look at the Jacobian we will see that so each of these equations can be solved using method of characteristics after these are first order linear partial differential equations because ABC are known functions of x and y only phi x and phi y so this is a linear equations of course how do you solve by characteristics method you have to write equations for characteristics. So characteristic ODE's are this so this equation look at particular third equation it says dz by dt is 0 it means any solution is constant along the base characteristics any of the each of the base characteristics you consider the function phi will be a constant on that. Now assume that phi of x y equal to k represents a one parameter family of solutions to this ODE this ODE is what it is a nonparametric expression of this base characteristic curves on differentiating this equation phi of x y x equal to k with respect to x in other words we are assuming that there is a solution hidden inside this of this ODE that is part of this assumption. So we get this by chain rule and from here we can solve minus phi x by phi y has certain expression why are we doing this because we want to show that the phi and psi chosen the way that we have described will give a change of coordinates. So we need to find what is phi x by phi y we have an expression now we decided psi will solve the other second equation therefore the corresponding characteristics ODE is given by this. So any solution as before because of this relation it is constant along base characteristics and what is a nonparametric expression of base characteristics that is given by dy by dx equal to this b plus root b square minus ac by a. So this is a present base characteristic curves as before differentiating psi of x y x equal to k we end up with an expression for psi x by psi y equal to this. Now b square minus ac is positive therefore this determinant is nonzero. So what is this determinant if you expand and because this expression that we have derived for phi x by phi y and psi x by psi y I will just bring it back see here b square minus ac nonzero if this is 0 for example it is just b by a for both the cases then the Jacobin will be 0. So since b square minus ac is nonzero this slope is different from the other one where we had a minus sign here with a minus sign that was for phi x by phi y. So go back and see that slide that is a reason therefore you have a change of coordinates because of the inverse function theorem. I am not writing all the details here after applying what do you get from inverse function theorem and formally conclude I am not doing that because now we are experienced with writing such conclusions because in first order PTE we dealt with them very clearly. Now since the transformed equation remains hyperbolic we observed if you set a and c are 0 b has to be nonzero dividing the transformed equation by b gives an equation that you wanted. We make one more change of variables what is the equation that we got now w xi eta plus something into w xi plus something into w eta plus something into w plus some function of xi eta equal 0. Now we are going to do another change of coordinates we write x dash and y dash as the function of xi eta which is xi plus eta and this is a function of xi eta which is xi minus eta when we do this change of variables we get a new transformed equation for v of x dash y dash if you observe v y dash y dash minus v x dash x dash this precisely how the wave equation look like utt minus uxx this rest of the part features only first order derivatives or no derivatives. The second order derivatives involve only in these two terms and that part looks like the one in the wave equation. So both the equations which we got which features only the mixed partial derivative w xi eta or the equation which we obtained here both are admitted as canonical forms for hyperbolic equation. Of course actually somebody has defined a canonical form with some other definition but we need not give value to such definition because after all just naming what is admitted as canonical form what is not admitted as canonical form it depends on what you idolize on right what you model what you like. So we will admit both of them as canonical forms no problem we do not lose anything. Let us look at an example I have already called this PDE as PDE dot hyperbolic instead of calling with some number and this equation is nothing very special equation it is just some second order PDE and therefore I call this already PDE hyperbolic because it turns out the equation is hyperbolic somewhere. So this is just name do not give too much value to this name this may be I could have called just PDE also okay. So for this equation what is a x square what is b minus x y what is c minus 3 y square. So what is b square minus c 4 x square y square it is always positive whenever x y is nonzero because it is 2 x y whole square that is always positive when x y is nonzero. Therefore except the places where x y is 0 this equation is hyperbolic and x y equal to 0 if and only if either x is 0 or y is 0 that means either you are on x axis or on y axis. Outside this which means in each of the 4 quadrants which do not include the axis the equation is hyperbolic. When such thing happens yeah just note that on the axis it is parabolic because b square minus ac is 0. Now we will transform the equation into canonical form in the first quadrant we need to do that because in first quadrant we know what is x and y x is positive y is positive because when we integrate certain ordinary differential equations longer them will invariably come and longer the mod x longer the mod y these kind of things and which can be uniquely fixed you know if you know where you are working that is the reason why we determine in first quadrant. Say in the other remaining quadrants also you can determine the canonical form. So in order to find a new coordinate system we need to solve these ODEs dy by dx equal to b plus or minus root b square minus ac by a we substitute the values of abc from our equation we get this. So in other words the two ODEs are this and this these are very simple ODEs you can solve them and solutions of this are given by y by x equal to constant solutions of the second ODE are given by x cube y equal to constant respectively. At this point do not worry that yes this is not defined when x is 0 of course x is never 0 we are in the first quadrant and which constants do not worry which constants because finally we see what we get. Look at this xi eta which is defined by this is a solution of the first ODE this is a solution of the second ODE and differentiate this equation connection between u and w because we need to plug in the values of u x, u y, u x, x, u y, y etc in the PDE. So we need to compute various derivatives please do the computations on your own pause the video compute the things derivatives. Now go back and substitute in the given PDE you get this of course this is not a good equation because you see that there is a x y here we need to eliminate that also express everything in terms of xi eta and then we have this formula right for x and y in terms of xi eta take this go ahead and substitute we get this. So on simplification we get this equation because we wanted w xi eta coefficient should be 1 so I divided everywhere so that means we need that xi and eta are never 0. Now we can write down in which domain we have determined this particular canonical form. So a method to reduce a second order linear PDE of hyperbolic type to its canonical form was presented of course it was also successfully implemented in example. In the next few lectures I already mentioned we are going to take up equations of parabolic type and of elliptic type and find canonical forms for them. Thank you.