 In this lecture, we are going to consider a partial differential equation which is studied very well in the literature. It is called Berger's equation and we consider certain initial value problems for Berger's equation. The Berger's equation is one of the simplest quasi-linear equations and it exhibits many features of solutions that is common to non-linear equations. Common in the sense, they occur very frequently for many non-linear equations. So, the non-linear effect in the Berger's equation, we already saw the Berger's equation, but we will again see the equation and then we will talk about it that. So, we are going to consider initial value problem. It is an equation which has one variable is called time variable. So, therefore, it is an initial value problem. We consider examples basically of the initial value problem, solve them explicitly. And then study, if I consider an initial condition with a monotonic function, what happens to the solution? So, question is why study Berger's equation? Why study this special equation? It is a simplest quasi-linear equation. Its solutions exhibit a typical behaviors of solutions to general quasi-linear and hence even non-linear equations. And the study tells us to be ready for certain surprises that we may find in studies of more general equations. So, we consider Cauchy data of this type Ux0 equal to Hx. If the variable y, it is a function of two variables, independent variables x and y. So, the variable y is interpreted as time. In fact, it is interpreted as time. So, this is nothing but initial value problem. That is why we write initial value problem for Berger's equation. In the examples that follow we use functions H of x which are monotonic functions, but they are of the following type, they are discontinuous or piecewise linear or smooth functions, differentiable functions. Of course, you may ask this question, why consider non-differentiable functions? Yes, our existence and uniqueness theorem is not applicable for such an H. However, we have to deal with such H, as they are physically relevant. We cannot always say that my theorem allows only smooth functions as the data, I will work with only that. No, these are also important, it is also important to study such functions H. Of course, in such cases, solutions needs to be interpreted differently, not the way that we have been talking about. So far, the idea of solution that we have introduced is what is called a classical solution. So, we have to now generalize the notion of solution. We have to define what do we mean by solution, so that whatever we do is meaningful. And computations are very easy for the kind of function that we propose to deal with. And formula for solutions within quotes can be derived. So, it helps in understanding properties of solutions very clearly. So, we find expressions or formula for solutions. We get formulas, basically some formula like f of x, something like that. Then we understand what is the domain of this f, for what x that f of x makes sense. These questions are to be asked later. So, first we find formulas and then ask, when are these formula actually give rise to solutions? Once again, observations made in these examples continue to hold with a smoother function. It is not that smooth function things are all right, no. But when you consider smooth functions, you cannot go from one constant to another constant in a smooth way. It takes some time where the function is not constant and computations are difficult and illustrating them by graphs is difficult. But we can be guaranteed assured that similar difficulties will be there even for such age. Only thing it is more complicated. So, why consider non-differential functions? Again, same thing. We are going to use non-differential functions even though they are bad. There is a general theory. This is some kind of assurance. There is a theory do not worry in which we can make sense of whatever we are doing. So, please go ahead and do, do not bother about whether it is correct or not. And that discussion is out of scope for this course. We will see a glimpse of it later on in the next lecture. References are books on partial differential equations by Evans, or Fuland Prasad and Renuka Ravindran. And there is a book by smaller shock waves and reaction diffusion equations. There are also many more beautiful books written on this kind of material. So, this is the Burgers equation. u y plus u u x equal to 0. We would love to write u t plus u x equal to 0. But since we are going to follow method of characteristics where t is used as a parameter running on the characteristic curve, we are not using the t here. We still write u y plus u u x equal to 0. And initial condition u x 0 equal to h x x in R y positive. So, let us parameterize the given Cauchy data. We need to solve the method of characteristics. Let us follow that. So, first is parameterization of the Cauchy data, x equal to s, y equal to 0, z equal to h s. Now, characteristic system of ODE, dx by dt is equal to a. In this case, that is a u x that is u which is z, dy by dt is 1, dz by dt is 0. We need to solve this set of ODE is, system of ODE is with this initial conditions. Observe that z does not depend on t. What does that mean? It means that solution is constant along each base characteristic curve. And the solution is satisfying initial conditions is easily given by this. First is z, dz by dt is 0. So, z is constant, at t equal to 0, it has to be h of s. So, that is h of s. And dy by dt equal to 1, therefore, y equal to t plus constant, but at 0 it should be 0, it is t. Now, when it comes to x, dx by dt is z. What is z? H s. Therefore, dx by dt equal to h s, if you solve you get this with this initial condition, very easy. Now, u equal to h of x minus y u, that is what we get using these three equations. Yeah, ideally we should solve for t and s in terms of x and y and then go and substitute here. So, h of somebody, right. So, if I show s is equal to x minus y u, we are done, which comes from here, s equal to x minus h of s is supposed to be a solution, u and t is the y. So, that is why you get this implicit expression u equal to h of x minus y u. Of course, when we admit this as an implicit solution provided you can solve for u as a function of x and y. So, note that this equation is meaningful, even if h is not differentiable, right. It is just asking that u equal to h of x minus y u, imagine h is continuous. Equation is meaningful. Now, question is whether you can solve u in terms of x and y is another question. We may be able to define a function u of x, y from this equation, defined on some subdomain of omega 2. What is omega 2? Recall, it is a projection of omega 3. What is omega 3? Omega 3 is where the coefficients of the Causillian equations are defined. Omega 2 is a projection of omega 3, 2, x, y plane. Solutions will be defined in some domain of, subdomain of omega 2. So, it is possible. And function thus obtained may turn out to be a solution, may turn out to be differentiable on a further subdomain. Therefore, it will be a solution to Berger's equation. So, therefore, given this possibilities what we do in the examples is we arrive at this equation, somehow solve for u and then ask on what domain it is differentiable, catch hold of that domain and say this function defined on this domain is a solution of the Berger's equation. That is what we are going to see in the examples later on. Now, for each fixed S, what does this represent? This is the base characteristics x equal to h s t plus s comma y equal to t. As t varies, s is fixed, what is this set? It is a straight line passing through the point as 0. This is the point s comma 0. So, it is a straight line passing through as 0, this point with slope 1 by h s. Therefore, what we observe is that it is possible in principle for example, like that these some at the next point if the slope is like that, so that equation straight line goes like this, this is not correct. So, at this point equation goes like that, it is fine they do not touch here, but on the other hand if you have like this and at this point slope is like that. So, they will intersect. So, base characteristics can intersect, it depends on h s, we will see that. So, this base characteristics we use a notation l s, l s is a straight line, alpha line s for passing through this point as 0 and this is the equation slope clearly 1 by h s. If you take two different s 1 and s 2 and take base characteristics through those two points, do the intersect we already saw the picture. This question is important, why are we asking this question? This question is important as we anticipate a potential problem. Recall that any solution to Berger s equation is constant along each base characteristic curve. For example, we had this and we had a base characteristic curve like that and we had a base characteristic like that. So, this point is s 1, 0, this point is s 2, 0. What we saw is any solution of the Berger s equation has to be constant on each base characteristic curve and what is the value here? It is h s 1 and here sorry this is h s 2, this is h s 1. So, therefore on the green line it is h s 1. On the blue line it is h s 2. What will it be at this point of intersection? Problem. See base characteristics they carry information from the gamma 2, this is gamma 2. It carries the information which is given namely u must be equal to h x that is the information given on this x axis that is being carried forward by these lines which are characteristic curves. So, they are carriers of information from gamma 2. So, on l s 1, u is h s 1, on l s 2, u is h s 2. What if they are not same? It means conflicting information is reaching at the points of intersection of the base characteristic curves. What will happen there? Solution becomes multi valued. If at all you want to say solution. So, there is a trouble at those points. That is why this question is important. Now where do they intersect? l s 1 and l s 2 we have two equations straight line equations. When you write down if they intersect at a point x 0, y 0 that should lie on both l s 1 and l s 2. This is the equation for l s 1 first line. Second one is a equation for the l s 2 and it should be satisfied. That means this system should have a solution for x 0, y 0. It is a non-homogeneous system. So, if the determinant is non-zero, definitely you have a system. You have a solution for x 0, y 0 and determinant is h s 1 minus h s 2. So, if h s 1 equal to h s 2, then there are parallel lines. They do not intersect therefore no problem. But if they are not equal then the system has a solution. They intersect at this point leading to an ambiguity in the value of the u at that point x 0, y 0 because it is both h s 1 and h s 2 and they are not equal. So, there is a problem. There is ambiguity. Should it be h s 1 or should it be h s 2? Now let us look at the transversality condition j t s, which is dou x y by dou t s is this. So, this turns out to be minus of 1 plus h dash s into t. From our existence and uniqueness theorem for Cauchy problems for quasi-linear equations, if this is non-zero, no problem. If this is 0, troubles to be expected. Where is it 0? Precisely when h prime s t plus 1 equal to 0. That is from our experience, recall the discussion during the proof of existence uniqueness theorem, we expect troubles at points where this happens. We will get back to this discussion later, later on. So, j of 0 s 0, this is actually where what we use to apply our theorem because we do not know at a general t. So, t equal to 0 only we have the information. But in the last slide, we have all the information. We have explicitly solved everything. That is why we could write j t s and we know this expression. But otherwise we do not know the particular the second column at arbitrary t only at t equal to 0. So, j of 0 s 0 is minus 1, it is never 0. So, it just means that there is a existence uniqueness at every point which is nearby at some points nearby the datum curve. This is the solution. Solution exists. Good. Once again as before the expression is meaningful even if h is not differentiable. But this may not represent a solution. The way a solution is defined, there could be problems. We go ahead without worrying about this for reasons explained at the beginning of this lecture. Let us look at specific examples. So, we are going to consider 4 examples of Cauchy problems for Burgers equation. Initial profiles, the function h of x, it is graph that can be called as initial profile. Those are monotonic functions that are piecewise linear or piecewise constant. Note that such functions are differentiable with exception of a few points. Monotonic functions are very close to being differentiable. So, they are differentiable almost everywhere. And explicit expression for u can be obtained in these examples. And the function exhibits the effects of nonlinear nature of Burgers equation and the monotonicity of the initial profile both. So, these examples highlight the role played by the nonlinearity of a partial differential equation in solutions to Cauchy problems. Now, in this example, method of characteristics fails to determine a solution in some region of the upper half plane, which means it is not global with respect to domain. So, consider Burgers equation with Cauchy data minus 1 for negative x and 1 from x equal to 0 onwards. LS we already observed the family of base characteristics are equations with slope 1 by hs. So, for s less than 0, LS has slope minus 1 and u is minus 1 on that because the data carried forward is minus 1. For s greater than or equal to 0, slope is 1 because hs is 1, 1 by hs is 1, hs is 1 therefore, this is 1 slope 1 and information it takes is 1. So, solution will be 1. Let us draw a picture. So, this is the picture. So, here we had s negative. This is for s greater than or equal to 0. These are the base characteristic curves. These are the base characteristic curves corresponding to positive s. So, these are all lines parallel to y equal to x. These are lines parallel to y equal to minus x. Now, what happens in the in between this region? In this region no base characteristic curves. So, no information is being passed from the initial data that is the gamma 2 into this region which is in the V shaped region. Dots means this not included because h of x was like that, right? h of x was minus 1 if x is less than 0. That is the reason why this 0 line is not included x equal to minus 1. So, in the V shaped region no base characteristics pass. Therefore, no information on the solution from x axis reaches there via base characteristic curves. This is happening because the jump in hx because suddenly shift from minus 1 to 1. Therefore, the slopes also across s equal to 0 that is what results in the V shaped region. Now, u equal to minus 1 is a solution defined on this domain. You can see in the picture 1 is a solution on this domain. So, what to do in the V shaped region? That is a natural question. What happens in the V shaped region we saw via base characteristic curves. But can you define a solution in the V shaped region? This is a question that we ask. Yes, something can be done so that we can define a solution. If you observe the solution, I have put it in quotes. Means some generalized notion of solution. But this discussion is out of scope for this course. So, I look up the references that I mentioned earlier whenever you are interested in this question. So, if we choose a h that goes from minus 1 to 1 in a continuous manner, then V shaped region would not be there, right? We observed that that is because a jump sudden jump from minus 1 to 1 we had this problem. So, think about this. Now, let us look at the second example. Here what happens is solution becomes multivalued in some places. So, h is now 1 minus 1. It is a decreasing function. It is a monotonic function but decreasing. Equation for the base characteristics is the same still. So, s less than 0, ls has slope 1 and it carries solution will be 1 there. And for s greater than or equal to 0, h is minus 1 therefore slope is minus 1 and solution will be minus 1. Let us look at the picture here. But these families intersect in this V shaped region. In this region, maybe this line is excluded but here base characteristics are intersecting and they carry 2 different informations. So, if you want to say the solution, at this point it is both 1 and minus 1 if you want to say that. So, you have to change your notion of solution. But we see that conflicting information is reaching there. So, there is some problem. l minus 3 and l 4 for example, l minus 3 is this line and l 4 is this line. So, they intersect here. At this point it is both 1 because of this and minus 1 because of this. Now, we can write down where 1 is a solution and minus 1 is a solution. Once again there is a problem in the V shaped region. The answer is same as given before. Something can be done which is out of scope. You can read the books that I have suggested. Did you notice that you are thinking of global solutions by asking this question. Once you get some solution in the V shaped region, you have the solution in the entire upper half plane. So, it is there at the back of our mind. Let us look at the third example. Here initial profile is very nice initial profile. It is a linear polynomial 1 minus x. No jumps, nothing. Continuously differentiable c infinity. It is analytic polynomial after all. So, let us substitute and get the expressions for u. We get u x y equal to 1 minus x by 1 minus y. Obviously, y should not be equal to 1. So, it means the line y equal to 1 u is not defined. So, it is defined on unit of 2 disjoint sets. 1 above the line y equal to 1, 1 below the line y equal to 1. U solves Berger's equation on each of these sets. Of the 2, only this one below the line y equal to 1 is in touch with x axis which is where that is a gamma 2. That is where initial data is prescribed. Therefore, that is a solution with this domain. So, this is a picture. Any 2 distinct base characteristic curves intersect at this point 1, 1 and solution becomes multivalued at 1, 1. Are there regions in the upper half plane where a solution is not determined apart from the line y equal to 1? It looks like yes, because there seems to be no characteristics passing through this, but that is not the problem because as you see the line from here is going like this. So, if you want to go like this, it might have gone out of the picture that is why you are not seeing. So, characteristics do fill up all these points except these dotted points base characteristics go through each and every point. Now, this is the most complicated example. Why? Because it is mixing all the 3 things 1, 1 minus x and 0. So, it is a piecewise constant and in between linear we will see what is going to happen. There are 3 distinct families of base characteristic curves depending on yes. Earlier we had only s less than 0 as greater than 0. Now, what will we have? We will have s less than or equal to 0, 1 case between 0 and 1 and bigger than or equal to 1. Let us call give names. So, it will be easy for us to refer to. F1 family corresponds to s less than or equal to 0. These are lines parallel to y equal to x because the initial condition is 1. So, therefore slope is 1 and therefore the value that solution takes along them will be 1. F2 family is that 1 minus x. So, solution will be like 1 minus x by 1 minus y. F3 family is where solution will be 0 and slope will be like 1 by 0. So, they are lines parallel to x equal to 1. Yes, yes. LS has slope changing from 1 to infinity whenever s is between 0 and 1 as s goes from 0 to 1. For s greater than or equal to 1, LS has infinite slope as we saw these are lines parallel to x equal to 1. So, question is do base characteristics intersect. So, we have to analyze cases whether family any 2 members of family f1 intersect, f1 and f2 intersect, f1, f3 intersect, any 2 members of family of f2 intersect or some member of f2 intersect, some family some member of f3. These are the various cases we have to analyze. So, take s1 and s less than s2 do the base characteristics LS1, LS2 intersect. So, this is the picture. Here these are all lines parallel to y equal to x. Here these are the lines parallel to y axis or x equal to 1 and in this region which is identified here in this region which extends base characteristics intersect solution becomes multi valued. So, here u is equal to 1 because my initial condition h of x is equal to 1 here. Here h of x equal to 0 for x greater than or equal to 1. Therefore, the solution is 0. Here the solution is 1 because of this. Here it was polynomial 1 minus x. Therefore, the solution will turn out to be 1 minus x by 1 minus 5. Now, at what point they intersect it is algebra. I will skip the algebra and they are all intersect here at this point. If you start here do not intersect. Now, the question to be asked is at this height do they intersect? Yes, they intersect because at these points. At this height do they intersect? Which height? This height? No, no two base characteristics intersect. Then I raise like you know high jump I raise the bar a bit and I ask is this the place? No. It turns out this is the place where base characteristics start meeting that is basically this family is what is responsible for that all of them are meeting at this. This is the first time if you why is called time t time the first instance where some trouble starts brewing is at t equal to 1. So, this is often called breaking time for solution. So, two members of F1 family clearly do not intersect because all of them are parallel lines. They never intersect. Similarly, two members of F3 they do not intersect. But two members of F2 they always intersect at this point. And then you can ask when is this family intersect this that is once again at this point because this line is going right. And here the smallest time would be this. So, please compute this. So, whenever base characteristics intersect then thing of interest is always what is the first time at which they start intersecting until that time they should not be intersecting. So, this is just algebra. So, I just skip but I just keep it on the screen for some time so that you can do the computation. As S1 varies in minus infinity 0 means F1 family. This is F2 family. S2 is in 0 comma 1 is F2 family. It is 1. So, I am showing this once again through picture. Now, F1 family and F3 family where do they intersect at 1, 1. It means some member of the family of F1 meets some member of the family F3 at the point 1, 1. Now, F2 F3 once again the point 1, 1. Now, both of them are from F3 family. They do not intersect because they are parallel lines. So, no two base characteristics intersect in the open region bounded by x axis and the line y equal to 1. We have observed this. Solution is uniquely determined in that region and is given by this formula. Observe that u is not a differential function on the line segment x equal to y and x equal to 1. Let us summarize. So, we discussed four initial value problems for Berger's equation. We understood that a solution may not be determined on the entire region omega 2 where the PDE is posed due to base characteristics not filling up the entire region. This was the case in one example or base characteristics intersecting with each other. This is true in the other two examples and carrying in conflicting information possibly. So, that is a problem. So, now question is how to overcome these obstacles of the type that we discussed above and have a solution. That means we have to define a new notion of solution. Wherever the PDE is defined we want that. In the context of Berger's equation, let us limit ourselves to the context of Berger's equation. That will be discussed in the next lecture where we will be worried only about how to give a meaningful solution or how to make sense of these functions which we obtained here as solutions in a some kind of generalized sense that we will be discussed in the next lecture. Thank you.