 Welcome. In this lecture, we are going to study a stronger version of the weak maximum principle that we considered in the last lectures. It is called strong maximum principle. We will also be discussing what is known as Dirichlet principle. Dirichlet principle roughly speaking it says that solving a Dirichlet boundary value problem is same as solving a minimization problem for a functional. Let us get into the lecture. The outline is consisting of two points, strong maximum principle and then we move on to discuss Dirichlet principle. So, let omega be a bounded domain in R2 and you be a harmonic function in omega. The weak maximum principle asserted that the maximum value on omega closure is attained on the boundary of omega. Of course, it never told where else the maximum is attained. In particular whether maximum is attained in the domain omega or not, it did not say. Strong maximum principle that we are going to prove says that if such a maximum is also attained inside omega then the harmonic function must be a constant function. So, let us state strong maximum principle as a theorem. Let omega inside R2 be a domain. Domain need not be bounded. Let u from omega to R be a harmonic function. If u attains its maximum in omega then u is constant. Notice in the hypothesis of strong maximum principle we are not assuming that omega is a bounded domain. Therefore, even if you have a continuous function which is even continuous on the closure of omega the maximum or minimum may not make sense. Therefore, strong maximum principle does not talk about that. On the other hand what does it talk about is if there is a maximum and that maximum is attained in omega then the harmonic function has to be a constant function that is what the strong maximum principle asserts. So, what is the idea of the proof? Assume that u attains the maximum value denoted by m. In other words there is a hidden assumption in the background that is supremum is indeed meaningful and that supremum is attained at some point in omega. That when it is attained the supremum is called maximum and let us denote it by capital M at some point P naught in omega. So, P naught is a point in omega where u attains the maximum value namely the m. Let P be any other point in omega. We will show that u of P equal to m. So, what does that mean? u of P equal to m for every P in omega that means u is a constant function. How do we show this? Step 1, u is locally constant near points of the maximum. We are assuming that the maximum is attained at P0. Therefore, what is the meaning of this sentence? u is locally constant near points of maximum. There is a disk around P0 on which u is constant and that constant is m. Step 2, continuation argument. We have shown in step 1 that u is constant in a disk around P0 but I want to show that u of P is also m. So, natural idea is to go from P0 to P using a curve and show all along the curve u is the constant m. Then it follows that u of P equal to m. We will see how this idea is implemented in the step 2. So, connect P0 and P by a curve. Try to continue the idea in step 1 from P0 till P. So, let us move to step 1. We are going to show that u is locally constant near points of maximum. I already mentioned how we should read this kind of words locally. So, this u is locally constant near points of maximum. It means take any point of maximum, then there is a disk around that on which u is constant. That is the meaning of locally constant. So, let u of P0 be m as we assumed where m is the maximum of u over omega. Let R be such that the closed disk d P0, r is contained in omega. Recall this notation. We use this closed brackets here, these brackets, square brackets to denote the closed balls or closed disks. If there is a point q in this disk P0, r where u is not m. In other words, if u is not m, m being the maximum, you will be strictly less than m. So, suppose this happens. Then by continuity of u, there is a disk around q, a closed disk around q on which the function remains less than m. This follows by continuity. So, in view of the mean value property, we know that m is given by u of P0. Now, u of P0 is given an average or the mean over this disk d of P0, r. This is the area of the disk pi r square. Integral of u over the disk divided by the area of the disk. So, therefore, this is the average on the disk. That is equal to u of P0. We know this because of the mean value property. So, we have stated this already on the last slide. Now, we write d of P0, r as union of two things. One is d P0, r minus dq epsilon, union dq epsilon. Therefore, the integral becomes some of these two integrals. Now, I know that on this, it is strictly less than m, u is strictly less than m. On this, u is less than or equal to m. Therefore, we have a strict less than now because the first term is strictly less than and this is a non-zero quantity. So, if you multiply with a non-zero non-negative quantity with a strict inequality, strict inequality is respected. So, we have strict inequality here. I have replaced u with the bound for u which is strictly less than m on the disk and u is less than or equal to m anyway in omega. Now, if you see these two integrals add up to an integral on d P0, r. Therefore, this evaluates to m because m is a constant. It comes out. What you have is integral of d P0, r which will give you the area and anyway you have an area here. So, both get cancelled and you get m. So, what we have got is m less than m. It is not possible. We conclude that u is constant on the disk d P0, r. Why did we get this contradiction of m less than m? That is because we assume that there is a point q at which u is strictly less than m. That is the reason why we got m less than m here. Given any real number, it has to be equal to itself. It cannot be strictly less than or strictly greater because of the law of trichotomy in real numbers. So, this concludes the proof of step one where we have established that there is r such that u is constant on disk of radius r with center at P0. Let us move on to the step 2, the continuation argument. Assume P0 is a point of maximum. Let m be equal to your P0. We are going to prove that u is a constant function that takes the value m everywhere in omega. So, take a arbitrary point P in omega. We will show that u of P equal to m. So, let gamma be a smooth curve without self-intersections. These kinds of things we need for the technical things that will follow. Otherwise, simply speaking, take a curve which connects P and P0. Existence of such a curve essentially follows from the path connectedness of omega which in turn follows from omega being open and connected. So, once you have a curve, you can always have a smooth curve. So, take it as a fact or try to do the exercise if you have enough background. Try to do this as an exercise. So, since gamma is a compact set, it maintains a positive distance from boundary of omega when omega is a proper subset of R2. So, let us denote this distance by D gamma. If omega is Rd, then take D gamma to be any positive real number. So, in other words, we have P0 here. We take any point P and take a curve gamma. Now, if this is, imagine this is a bounded domain omega or domain is omega is like this. It has boundaries. Then what is the distance of this curve to this boundary of omega? That is a positive number. So, we have to see which is the closest point. Perhaps this is the closest point in this picture or maybe this. So, if you call D gamma as a distance, what does that would mean is that if you take at any point on this curve, take a circle of radius strictly less than D gamma, it will not intersect the boundary of omega. That is the reason why we are taking this. In other words, this ball of radius which is strictly less than D gamma is properly contained in omega. Of course, if omega is Rd, you can take any D gamma. You do not have to be careful at all. So, in either of these two situations, the disc of radius D gamma by 2 which is strictly less than D gamma with center at P0 denoted by D P0 D gamma by 2 lies in omega. Not only this, even the closed disc, the closed bracket P0 D gamma by 2 closed bracket that also lies in omega. So, by step one, we know that u is constant on this disc because P0 is a point of maximum. Therefore, u is locally constant, we proved in step one. Therefore, u is identically equal to m on this ball, on this disc. Actually, step one says that u is locally constant around points of maximum of u. But if you carefully observe step one, what we have actually proved is that whenever you find a disc with center at P0 where P0 is a point of maximum of u such that the closed disc is contained in omega, then u is identically equal to m on this disc. Now, for a point q in gamma, let D P0 q denote the distance from P0 to q along the curve gamma. So, let me just illustrate. Suppose we have point P0, P here and u take any point q. The usual distance is the Euclidean distance which is the length of this line. But what I am asking you to do is take the length along this curve along this curve. You know that if you move along curves and not along straight line, the distance will be more. So, go along this, that distance is called D of P0 comma q. So, we now take a point P1 on gamma which lies in this disc P0 D gamma by 2. That means, so this is P0, this is P and by step one, we have proved that on this disc whose radius is D gamma by 2, u is constant. Now, we plan to take a point P1 which is on this curve lies in this disc. But where will we take? Will we take here? Will we take here? Let us see more prescription. We are going to take on this curve. I want P1 to be on the curve gamma also. So, I will take P1 inside this here. This is my P1. So, P1 is inside this ball. P1 is here. Not only that, I want to maintain some distance D of P0 comma P1. I will take it to be D gamma by 4. Recall what is this D P0 P1? It is a distance along the curve D gamma by 4. Therefore, P1 will lie inside because along the curve I am taking a smaller distance D gamma by 4. Whereas, the disc is D gamma by 2. So, this point P1 cannot be outside. If a P1 satisfies this criterion that the distance along the curve is D gamma by 4, it is not outside this disc. Therefore, it will be inside. So, it is possible to find such a P1 which is in this disc as well as this criterion. I have already explained why is it possible to find such a point. And u of P1 equal to m because P1 is in the disc P0 D gamma by 2. Therefore, u of P1 is m. Now, repeating the above argument, get points P2 P3 and so on till you get a k in n such that P belongs to the disc with center Pk and radius D gamma by 2. So, in each of these steps when we try to find P2 P3 etcetera, we are insisting on this that P l of course belongs to the previous one P l minus 1 radius D gamma by 2. We are not compromising on the radius. It is always the same. And importantly, the distance between the center and the point we are choosing is D gamma by 4. The distance is fixed. That means we are definitely moving along this curve P1 is here. So, this distance along the curve is D gamma by 4 and it is here. So, these two distances are same. See, sometimes it may happen that you are moving from some point to another point in the step 1. Let us say x0, x1, x2 but then you may be never crossing this some point x star. It can happen that the steps are becoming smaller and smaller and you are getting accumulated somewhere. You are not crossing this x star. But every time if you move a fixed step like this, you will definitely exhaust the distance which you need to do is simply D of P0, P. That is the distance if you cross using these tiny steps D gamma by 4, how many number of times that is the thing really here. So, l is equal to 1 to k perhaps. Then definitely we will exceed this. So, somewhere before 4, the P will fall into one of these discs. This is extremely important. We will then have u of P equal to m because u at P k is m and u is constant on this disc. Therefore, at any point in this in particular at P which is in this disc, u will be m. This finishes the proof of the theorem. Another proof let us look at. Define a set S by set of all x in omega such that ux equal to m. S is a non-empty subset of omega because we are assuming that u takes the value m. What is m? m is the maximum value of u. u takes the value m at some point in omega that is assumption. Therefore, the set is non-empty. And it is a close subset of omega because u is a c2 function in particular continuous function. Continuous function equaling a constant will be a close set. You can also think like this. This is a continuous function. Constant function is also continuous. So, two functions are continuous. That set where they are equal is a close set. Or you can also look at ux minus m equal to 0. And set of all x where a continuous function takes the value 0 is a close set. Now, it is an open subset of omega. It is not clear just from this definition. Using continuity you can only prove it is closed. We have to use something extra that we know about u namely u is harmonic. Open set is exactly the step one in our proof where we have proved that if u takes the maximum value at some point, then there is a disc around that point where u takes constant value m. That is precisely the meaning of showing this set is open set. So, we have got a set which is non-empty open and closed. It is a subset of omega. What is omega? It is an open and connected set. There you have a subset which is non-empty and both open and closed. By connectedness of omega, such set has to be the full set. That is s equal to omega. This is another proof because we are using here omega is connected. The proof looks so simple because we are using the fact that if you are working in a connected set or a connected topological space, a subset which is both open and closed has to have only two choices. Either it is an empty set or the whole set. We are using that result here. That is why it is simple. And step one proof anyway, we have to supply here. So, essentially step two is removed and we are appealing to the connectedness. Of course, we have used the connectedness in another format in the other proof also. A remark on the proof of strong maximum principle, both the proofs use the mean value property of u, namely in step one. Mean value property is an exclusive property of harmonic functions. Recall, we have not only shown that every harmonic function has mean value property, but also every continuous function which has mean value property is harmonic. So, almost mean value property is exclusively property for the harmonic functions. Harmonic function means solutions of Laplacian u equal to 0. But strong maximum principle holds for a larger class of elliptic operators for which mean value property may not hold because general elliptic operator need not be just Laplacian all the time. So, there are operators which are more general than Laplacian for which also the strong maximum principle holds. And the proof uses what is known as half's lemma, half, lemma of half. Most of the text dealing with general elliptic operators have the necessary details you may cancel them if you are interested. But in this course, we will not go beyond Laplace equation as I have pointed out at the beginning. Strong maximum principle asserts that if a harmonic function attains its maximum in a domain omega bounded or otherwise, then it is necessarily a constant function. Note that strong maximum principle does not comment on existence of a maximum value for harmonic functions. It does not comment about location at which supremum is attained if it exists. Let us look at an example. Let omega be the domain exterior to the unit disk. It means my domain is R2 minus the unit disk. That is origin radius 1, my omega is this. This function u xy equal log s square plus y square is a harmonic function on omega. This can be easily checked. u has neither a maximum value nor a minimum value in omega. Let us look at Dirichlet principle. Before that, let us recall some facts about system of linear equations. So, let a be symmetric positive definite matrix. Let small b be a vector in Rd, then the following statements are equivalent. x belongs to Rd is a solution to the linear system Ax equal to b. That is same as saying x is the minimizer of this functional g of y half y transpose Ay minus y transpose b. Dirichlet principle is an analogous result in the context of Dirichlet boundary value problems. This is a very useful idea. Demonstrating its utility is beyond the scope of this course. Nevertheless, let me mention couple of points. We have many methods to solve the system of linear equations Ax equal to b. For example, we have what is known as direct methods which will give us exact solutions like Gaussian elimination method and some modifications of that. Exact methods are good, but when this A is a big size matrix, matrix of big size that is D is very big, then it is not profitable. In fact, lot of errors might get enhanced in the method of solution. And people have found that conjugate gradient method is one of the very useful methods which is based on minimizing this functional. So, Dirichlet principle is analogous to this result and if you understand the utility of this result, it is easier to understand how this will also be useful. In fact, these kind of ideas are used in establishing solutions to elliptic equations and that method is also called calculus of variations. And the method itself is called, the first method is called the direct principle of calculus of variations where they will look at a minimizing sequence that is a sequence. For example, in this context sequence of vector yn such that j of yn converges to infimum of this functional and show that yn converges to some y and j of y is actually the minimum of the functional. So, that is the general idea in the direct method in calculus of variations. What is Dirichlet principle? Let omega be a bounded domain in R2 with smooth boundary. Let F be continuous on omega bar, G belongs to C of boundary of omega and U is C2 of omega bar. Then the following statements are equivalent. The function U is a solution to the Dirichlet boundary value problem minus Laplace in U equal to F in omega and U equal to G on the boundary of omega. And that is same as saying U is a minimizer of the functional J defined by this formula below and defined for V which is in C2 omega bar such that V equal to G on boundary of omega. So, minimizer in this set because if U is a solution to Dirichlet boundary value problem, U is in this set. U equal to G on boundary of omega, U is C2 of omega bar by our assumption. So, for a C2 omega bar function these two statements are equivalent. If you know that U is a solution to the Dirichlet boundary value problem, you can prove that it is a minimizer of this functional and so conversely if U is a minimizer of this functional, then it actually solves the Dirichlet boundary value problem. So, let us denote this set by S. So, let us move to proof of 1 implies 2. Let U be a solution to the Dirichlet BVP and let V be an element of S, multiply the equation minus Laplace in U equal to F with U minus V and then integrate on omega by this we get this. Then integrate by parts on the LHS that means Laplace in U becomes grade U and you will get a gradient here which is here gradient of U minus V is grad U minus grad V. No boundary terms from integration by parts because U equal to V on the boundary. So, rearranging the terms in this equation we get this. So, bring this term to this side take this term here to that side. Now, look at this inequality integral grad U dot grad V is less than or equal to integral norm grad U into norm grad V simply because A dot B is less than or equal to norm A into norm B this is a Euclidean norm of a vector. So, it is grad U of X grad B of X is less than or equal to norm grad U of X into norm grad V of X. Now, here this integral is less than or equal to this integral after that this is called first term as A second term as B is this is A into B that is less than or equal to A square plus B square by 2 I have used that. So, we get this. So, using the inequalities on the last slide we get that integral over omega of norm grad U of X square dx minus integral over omega of f U is less than or equal to this minus integral over omega of f V. So, rearranging terms we get this all the terms featuring U on one side and V on the other side but what is this? This is nothing but J U is less than or equal to J V this is the definition of J U this is the definition of J V thus U is a minimizer or the functional J over the set S this completes the proof of 1 implies 2. Let us look at proof of 2 implies 1. So, let V be a C 0 infinity function defined on omega and T be a real number then we have J of U plus T V you plug into the formula of J you get this expression and rewriting this what we get is J of U plus T V minus J of U take it this to the left hand side this remains as it is RHS. So, since the functional J achieves its minimum at U that is the hypothesis in 2 the function H from R to R defined by H of T equal to J of U plus T V minus J of U that achieves minimum at T equal to 0 thus H prime of 0 is 0 which will give us this relation integrating by parts on the LHS will give us minus Laplacian into V so integral minus Laplacian U into V equal to integral F V and this is true for every V which is C 0 infinity of omega. Therefore, minus Laplacian U of X equal to F of X note that U satisfies the boundary condition as U belongs to the set S the set S in the definition itself includes that U equal to G on the boundary of omega. So, here we need not work with C 0 infinity of omega we can as well work with C 0 2 of omega that is C 2 functions with compact support in omega. If this equality holds for every V which is C 2 0 of omega then you have that minus Laplacian U equal to F holds at every point of omega. So, this proves that U is a solution to the BVP thus completing the proof of 2 implies 1. So, let us summarize what is done in this lecture we have proved a strong maximum principle and we have established the equivalence of Dirichlet boundary value problem and a minimization problem. Thank you.