 Welcome to this lecture on weak maximum principle and its applications. The outline for today's lecture is first we introduce weak maximum principle and prove it and then we look at some consequences of weak maximum principle. We will discuss weak maximum principle only for domains in R2 for Rn it is similar. So, let omega inside R2 be a bounded domain. Let you be a continuous function defined on omega, such that you can be extended to the closure of omega as a continuous function. That means you make sense for points on the boundary of omega and that too in a continuous manner. So, collection of such functions is denoted by c of omega bar or omega closure. When omega is bounded every function in c of omega bar attains both its maximum and minimum values because omega bar is a compact set continuous function and a compact set attains both minimum and maximum values. But somewhere in omega bar the location is not known but somewhere in omega bar definitely both maximum and minimum values are attained. Now we have a question for any function u that is for any given function u in c of omega bar can we guarantee that the maximum is always attained in the domain omega or maximum is always attained on boundary of omega? So, can we answer this question for any arbitrary continuous function on omega closure? Answer no we cannot say this because there are functions which attain maximum in omega only and not on the boundary and there are others which attain maximum only on the boundary of omega not in omega. However, if u is a harmonic function that is u is a solution of Laplace in u equal to 0 then u attains its maximum value definitely on boundary of omega. This is the content of weak maximum principle. So, let us describe what is weak maximum principle. Let us state it as a theorem proof is very simple. Let omega inside R to be a bounded domain. Let u be in c 2 of omega intersection c of omega bar and u is a harmonic function in omega. Then the maximum value of u in omega bar is achieved on the boundary omega. We have a question here. Let m be the maximum value of u on omega bar. What do you understand from the conclusion of this weak maximum principle which is here? Choose all options that apply. It is a MSQ question which you are already very familiar with that means more than one answer can be correct. First option A is there is a point on boundary of omega at which u takes the value m. Second option there is no point in omega at which u takes the value m. Definitely the conclusion is saying that maximum value of u in omega bar is achieved on the boundary. That means there is a point on the boundary at which u takes the value m. So, A is correct. What about B? B is correct or not? B is true or false? When we write statements in English we have to be very careful when the maximum value of u is achieved on the boundary. So, if you put extra emphasis on this word ease, it is ease achieved on the boundary that gives an impression that it is not achieved in omega whereas this statement does not say. So, you are likely to tick B also but that is not true. There are harmonic functions which attain maximum inside omega as well as on the boundary of omega as guaranteed by this theorem. So, you have to be very careful with the conclusions which are written in a language like this. You should not give extra emphasis on these kind of words in the language. Be careful. We have come across such statements in the first order PDEs also. Proof of weak maximum principle. Step one, if a function u has a local maximum at a point in omega, omega is an open set then the second order partial derivatives are always less than or equal to 0. So, dou 2 u by dou x square at the point p as well as dou 2 u by dou y square at the point p is less than or equal to 0. Both of them are less than or equal to 0. As a consequence Laplacian u at the point p is less than or equal to 0 because Laplacian u after all by definition is u x x at p plus u y y at p. Both of them are less than or equal to 0. Therefore, Laplacian u at p is less than or equal to 0. Thus, if v is a function such that Laplacian v is positive in omega that means this kind of condition cannot be satisfied by v. Laplacian v at p is always greater than 0. Therefore, it is never less than or equal to 0. Therefore, we can conclude that maximum value of v on omega bar is never achieved in omega. It is achieved in boundary of omega because it is definitely achieved somewhere in omega closure. It is not achieved in omega. Therefore, it has to attain on the boundary of omega. This is the main idea in the proof of weak maximum principle. What we are given is a harmonic function. So, Laplacian u is equal to 0. Therefore, this idea cannot be implemented straight away. So, we construct a v such that Laplacian v is positive and hence v has a maximum on the boundary of omega and using that information, we show that u also has maximum on the boundary of omega. So, define the function v epsilon by u of x y plus epsilon times x square plus y square. So, when epsilon equal to 0, you are at u of x y. So, this is you can think this is a perturbation of this function. But what kind of perturbation? I am going to consider epsilon positive x square plus y square is always greater than or equal to 0. Therefore, this term is always greater than or equal to 0. In particular, v epsilon of x y is always greater than or equal to u of x y. We will use this observation later. And v epsilon is a C2 function in the domain omega because u is a C2 function. What we are adding is definitely a C2 function. For the same reason, v epsilon is also C of omega bar. The reason being that u is in C of omega bar, this function is continuous everywhere in R2 in particular on omega bar. So, Laplacian v epsilon is positive because what is Laplacian v epsilon? It is Laplacian u plus Laplacian of this. What is the Laplacian of this? It is epsilon times Laplacian of this. So, v epsilon having this property that Laplacian v epsilon is positive in omega, it attains its maximum only on the boundary of omega. Let us denote by m maximum of u on the boundary and l to be maximum of x square plus y square on the boundary. So, we have v epsilon of x y less than or equal to m plus epsilon l for every x y in omega, for every x y in omega. Why is that? Because m plus epsilon l is precisely the maximum of v epsilon and maximum of v epsilon is maximum of u plus maximum of epsilon times maximum of x square plus y square. That is the reason because we have observed that v epsilon maximum is attained only on the boundary. So, therefore the values of v epsilon for x y in omega is always less than or equal to maximum of v epsilon on the boundary of omega which is less than or equal to m plus epsilon l. Since u of x y is less than or equal to v epsilon of x y because this what we are adding is a non-negative quantity. So, epsilon is positive. Therefore, we have this inequality and v epsilon of x y we know is less than or equal to m plus epsilon l. Therefore, u is less than or equal to m plus epsilon l for every x y in omega. Note that the last inequality holds for every epsilon positive and this side there is no epsilon. So, as epsilon goes to 0 what we get is u of x y is less than or equal to m and that is what we want to show. What is m? It is a maximum of u on the boundary so what we have shown by this inequality is the value of u at any point in omega is less than or equal to the maximum value of u on the boundary. In other words maximum is definitely achieved on the boundary that completing the proof of weak maximum principle. A quick corollary of weak maximum principle is what is called weak minimum principle. Let omega be a bounded domain in R 2. Let u be c 2 of omega intersection c of omega bar and harmonic function. Then the minimum value of u in omega bar is achieved on the boundary. Proof is very simple. If u is a harmonic function minus u is also a harmonic function. So, let us consider v equal to minus u. Then v is a harmonic function. Apply the weak maximum principle to the harmonic function v and conclude. So, drawing the conclusion is left an exercise to you. It is a very simple thing. So, let us understand the conclusion of weak maximum principle. What does it say and what it does not say? So, weak maximum principle is proved for bounded domains. That is to be remembered. Look at this function u of x y equal to x. Of course, it is a harmonic function Laplace in your view is 0 throughout R 2. But I am now considering this only on this domain R 2 from which I have removed the closed disk with center at the origin and radius 1. So, this is not a bounded domain. In fact, is what is called an exterior domain as we discussed earlier. Where does u achieve its maximum value? Of course, u does not have a maximum at all. The weak maximum principle is silent on whether the harmonic function will take or will not take the maximum value in the domain. It is always saying maximum value is taken on the boundary but never says a sentence about what happens in omega. Let us look at this function u of x y equal to x square minus y square. This is a harmonic function this is a polynomial. So, this is also called harmonic polynomial sometimes. It is a harmonic function everywhere because Laplace in u will be 2 minus 2 that will be 0. So, Laplace in u is 0 everywhere but I consider on D 0 1 the disk of radius 1 with center at the origin. It attains its maximum only on the circle S 0 1. Why is that? If at all it attains maximum inside the disk inside the open disk we know that derivative of this function must be 0 gradient must be 0. What is the gradient of this function? It is 2x comma 2y. Where is it 0 at the origin? So, at the origin value is actually u of x y equal to 0 at the origin but 0 is not the maximum because clearly you can see if you are looking at the point u of 1 comma 0 that is actually 1 thus 0 is not a maximum value. Similarly, it is not a minimum value also. So, the only critical point that is where the gradient is 0 by definition in the disk is the origin at which u takes a 0 value 0 is not minimum value of u also. So, this is a function where maximum minimum is not attained inside the disk it is attained only on the boundary. On the other hand if you look at u x y equal to 1 it is a harmonic function it is a constant function. So, maximum and minimum both are 1 it is achieved everywhere in the domain as well as on the boundary. Let us look at some consequences of weak maximum principle. In lecture 6.1 uniqueness of solutions to Dirichlet boundary value problem was proved. In fact, we proved uniqueness for the Robin boundary value problem and Dirichlet boundary value problem turned out to be a special case. Uniqueness result can also be proved using weak maximum principle. In fact, weak maximum principle because it says maximum of u is attained on the boundary it compares two quantities maximum of u in omega is less than or equal to maximum of u on the boundary because it is in the form of inequality we get some estimates. So, weak maximum principle gives rise to a stability estimate we are going to see that. So, that is proving continuous dependence of solutions on the Dirichlet boundary data. Uniqueness is a simple consequence of the stability estimate. Let us state the stability estimate. Let omega inside r2 be a bounded domain and for i equal to 1 to 2 let ui be c2 of omega intersection c of omega bar solve the Dirichlet problem with right hand side same right hand side f and gi as the boundary data. Then the following stability estimate holds that is maximum of mod u1 of xy minus u2 of xy as xy vary in omega closure is less than or equal to maximum over boundary of omega of mod g1 xy minus g2 xy. If you want this is the distance between the Dirichlet data and this is the distance between the solutions in omega bar. So, define w equal to u1 minus u2 look at what problem w solves Laplacian w is 0 because Laplacian of u1 as well as Laplacian of u2 is f and w equal to g1 minus g2 on boundary of omega applying the weak maximum principle we get w of xy is less than or equal to maximum of the Dirichlet data the boundary data on the boundary because w is a harmonic function maximum is attained on the boundary therefore we have this inequality of course this is less than or equal to maximum boundary of omega mod g1 minus g2 because g1 minus g2 is always less than or equal to mod g1 minus g2. On noting that minus w also satisfies a BVP as above what does minus w satisfy Laplacian minus w is 0 and minus w equal to g2 minus g1 on boundary. So, we can apply weak maximum principle for minus w and what is the boundary data for minus w g2 minus g1. So, therefore weak max principle gives this now we know g2 minus g1 is less than or equal to mod g2 minus g1 which is same as mod g1 minus g2 therefore we have this. Now what do we have w of xy is less than or equal to certain quantity minus w of xy is also less than or equal to the same quantity for every xy in omega if a number and a negative of that number both of them are less than or equal to the same quantity it means the modulus of this number namely mod w of xy is less than or equal to that number. So, stability estimate follows from the last two inequalities. Now let us look at uniqueness of solutions to Dirichlet boundary value problem. Let omega be a bounded domain in R2 and consider this Dirichlet problem. We want to show uniqueness how do we show that we have to take g1 u2 satisfying this boundary value problem subtract look at the boundary value problem that is solved by the difference. If u1 and u2 are solutions to this problem u1 minus u2 let us call it w it satisfies Laplace in w equal to 0 and w equal to 0 on the boundary. So, from the stability estimate we get the uniqueness of solutions to Dirichlet boundary value problem. A remark on uniqueness result recall that the uniqueness result was already proved in lecture 6.1 recall that its proof required that the normal derivative of u is defined on the boundary and thus uniqueness result could only be proved for u in c1 of omega bar that is u in ct of omega intersection c1 of omega bar. Thanks to maximum principle uniqueness result is valid for any harmonic function which belongs to the space c of omega bar. In other words ct of omega intersection c of omega bar. Here I am mentioning only the smoothness what is required on the boundary. In the domain of course u is in ct of omega and uniqueness result does not hold on unbounded domains. Consider the Dirichlet boundary value problem posed on the upper half plane Laplace in u is 0 and u is 0 for all x in r. So, upper half plane the boundary is the x axis. So, this BVP has at least two solutions u1 of xy equal to xy xy Laplace in will be 0 and when I put y equal to 0 this is 0. So, it is a solution to this boundary value problem. Of course 0 is also a solution. So, we do not have uniqueness for Dirichlet boundary value problems on unbounded domains in general. Summary weak maximum principle was proved. We deduced continuous dependence on boundary data for Dirichlet boundary value problem from the weak maximum principle. Thank you.