 Welcome. In this lecture we are going to consider Dirichlet boundary value problems for Laplace equation. We will obtain a representative formula for a solution to the Dirichlet problem which is called Poisson's formula. It will be derived using Green's functions and towards the end of the lecture we will see certain applications of Poisson's formula. This lecture is as follows. We will recall certain things that we have done in lecture 6.2 namely fundamental solutions and their properties. Then we go on to define Green's function and we compute Green's functions via what is called a method of images for a ball in Rd when d is greater than or equal to 3. Of course ball is a disc when it is in R2. Then we write the Poisson's formula for the solution to Dirichlet boundary value problem on this domain ball or disc. As an application of Poisson's formula we will prove Harnak's inequality. A further application of Harnak's inequality is going to be Louisville's theorem. Let us recall from lecture 6.2. In lecture 6.2 fundamental solution for Laplace operator was defined. The fundamental solution for Laplacian is the function k from Rd cross Rd minus diagonal to R. We will define what is a diagonal once we see the formula for k. k of x xi equal to 1 by 2 pi log of norm x minus xi if d equal to 2 and if d is greater than or equal to 3 the formula for k x xi is 1 by omega d into 2 minus d into norm x minus xi power 2 minus d. As you see when x equal to xi both these functions have singularities therefore we have to remove that. So diagonal stands for the set x xi in Rd cross Rd such that x equal to xi and the following result was established. It says that Laplacian of k of x, xi equal to delta xi. For every xi fixed in Rd this Laplacian is in the x coordinates so Laplacian k of xi equal to delta xi. What it means we have explained that means actually this you can think this as a notation for the moment because we do not know what this is. In case you know this is exactly the same as what you know if you do not know this means this. For every phi which is c infinity compactly supported function in Rd phi of xi is given by integral over Rd of k x xi into Laplacian phi of x dx. Then we have stated this theorem on logarithmic potential it says that if you have a function in which is c2 and has compact support in R2 then if you define the logarithmic potential by this formula this u has a property that Laplacian u equal to f and u xi goes to infinity as norm xi goes to infinity this is a precise asymptotic as xi goes to infinity. Logarithmic potential is the only solution to Laplacian equal to f having this kind of asymptotic behavior. Then we have stated a theorem on Newtonian potential this is applicable for all d greater than or equal to 3. So if you have once again c2 function with compact support and define the Newtonian potential by this formula then the following assertions can be proved Laplacian u equal to f u goes to 0 as norm xi goes to infinity and this is the only solution to Laplacian u equal to f which is c2 and vanishes at infinity. Remark fundamental solutions help in solving Poisson s equation Laplacian u equal to f in Rd this is the essence of the theorems that we saw. So once the fundamental solution is known a single formula gives solution to Laplacian u equal to f it is logarithmic potential if it is d equal to 2 it is Newtonian potential if d is greater than or equal to 3 for any f and suitable. Question is there a function which behaves like fundamental solution does by knowing which any Dirichlet problem for Laplace equation can be solved. The answer is yes it is called Green s function understand the difference between Green s function and fundamental solution these two names are often blurred in the literature. So pay specific attention to the difference between these two. So definition of Green s function Green s function is defined as a solution to the boundary value problem Laplacian g equal to delta xi in omega g of xi equal to 0 on boundary of omega. Since the fundamental solution k already satisfies Laplacian k equal to delta xi in omega we look for g that is a Green s function having this form that g is equal to k minus phi now you know the equation for g you know the equation for k and also the condition for g on the boundary therefore you know what is the equation phi must satisfy. So phi solves Laplacian phi equal to 0 in omega and phi of xi equal to k of xi on the boundary of omega. So if you want to know the Green s function since we already know k what remains is to find is phi. The question is does Green s function always exists Laplacian operator is clearly the simplest elliptic operator that we can think of it is a non-trivial matter to show the existence of a Green s function. Leave alone finding explicit expressions for it for arbitrary domains omega. These are the two references I am going to give where you can find the discussion on Green s functions existence. One is a book by P. D. Lacks on functional analysis here you can find existence of Green s functions and arbitrary domains in R2. Then there is a paper on the existence of Green s function in proceedings of AMS by P. D. Lacks where he discusses Green s functions on arbitrary domains in Rd. For special domains like a ball or upper half space like the upper half plane and so on it is easy to construct Green s function. We will construct Green s function for a ball in Rd d greater than or equal to 2. So ball when d equal to 2 is called disc. So computing Green s functions we use what is called a method of images. So what is method of images? The method of images it is also called method of electrostatic images will be used to obtain Green s functions. The method prescribes that phi of x i is the potential due to an imaginary charge q placed at a point xi star which is not in the domain omega which is outside the domain omega and such that for every x on the boundary of omega the value of phi of x i coincides with the value of k x i which is created by the unit charge at xi. That is phi x i equal to k x i for all x in boundary of omega. So Green s function for ball in Rd d greater than or equal to 3 we will deal the case when d equal to 2 separately because of the way the formula for the fundamental solution looks is different. So method of images for d greater than or equal to 3 the potential due to a charge q at point xi star which is not in omega is given by exactly the fundamental solution. This is the charge q fundamental solution is for the charge 1 so now it is q now location is xi star thus we take phi of x i equal to this. In order to determine Green s function we need to find q and xi star so that phi x i equal to k x i for all x in boundary of omega. So using this constraint we have to find both q and xi star that is we need to find q and xi star such that this equation holds for every x such that norm x equal to R because the boundary of the ball of radius R is a sphere of radius R which is norm x equal to R. On simplification the above equation reduces to this because right away you can see that this cancels with this both are same factors and then you get q into norm x minus xi d minus 2 equal to norm x minus xi star power d minus 2 and take the power 1 by d minus 2 we get this. Squaring both sides of the last equation and rearranging the terms will give you this. At this point of time pause the video compute for yourself and make sure that you are getting this equation. Now assume since the left hand side of this equation is independent of x and right hand side depends on x both must be constant and that constant must be 0 please justify for yourself why should the constant be 0 that is this is 0 that is the first equation and Lh is 0 that is a second equation. Now the first equation holds for every x in s of 0 R the sphere therefore xi star minus q power 2 by d minus 2 xi that itself is 0. What it says is that this is some vector in Rd whose inner product with x is 0 for every x on the sphere of radius R if that happens then that vector has to be 0. So I have already given you the hint please work it out. Thus q and xi star satisfy these two equations. Now substitute the value of xi star from here into this at this point that will give us an equation like this. So on the last slide we obtained this equation now we need to solve for q there is no xi star here only q. So let us observe that if q is equal to 1 then xi star equal to xi and thus xi star belongs to the ball which is not allowed we want xi star outside the ball therefore q cannot be equal to 1. Thus for xi non-zero the above equation yields this how do I get this I just see that the unknown point is q agreed but then q power 2 by d minus 2 and its square is here q power 2 by d minus 2 whole square is precisely this. So it is a quadratic equation in q power 2 by d minus 2 think that is lambda and solve this quadratic equation you will get this. So summarizing the computations we have obtained q power 1 by d minus 2 equal to this in other words we have q and xi star is given in terms of q q is already known from here therefore we know q and xi star and this is nothing but r square by norm xi square into xi. So recall that phi of xi we started with this so therefore phi of xi equal to this I have just substituted what is q here and I have not yet substituted what is xi star the value of xi star inside this I still retain it as xi star xi star equal to r square by norm xi square into xi. So our search for Green s function proceeded by the decomposition formula g equal to k minus 5 therefore k is already known we have just found phi therefore g is known and that is given by this formula g of xi equal to this formula where xi star is this. Since norm x minus xi star you get this expression please compute this then you get that multiply norm xi this side then you have norm xi into norm x minus xi star equal to this numerator that as xi goes to 0 goes to r square therefore we can define g of x 0 to be this. Now let us turn our attention to finding Green s function for disk in R2. Once again method of images the potential due to a charge q at point xi star not in omega q by 2 pi log norm x minus xi star but we take phi of xi equal to q by 2 pi log norm x minus xi star as above plus a constant which I have written as c by 2 pi for convenience without this I have tried but I could not get anything so we have to add this constant then life is simpler this idea is taken from Wolver s partial differential equations book. So in order to determine Green s function we have to determine what q xi star and c such that phi equal to k on the boundary of the disk. So that is we need to find qc xi star such that this equation is satisfied for every x such that norm x equal to r on simplification exponentiate both sides you get this equation. So it is not clear how to solve the above equation for the unknowns qc and xi star a geometric idea helps here the geometric construction is as follows choose xi star as xi star equal to c xi that means it is along the line joining origin and xi so in the direction of xi. In such a way that for all x such that norm x equal to r the triangles formed by the vertices 0 x xi star and by the vertices 0 xi x are similar similarity requirement yields these equations. Let us try a circle just for explanation this is origin let us say this point is xi. So now we take a point outside the ball with what property is that you take any point x on the boundary we have this triangle this now choose a xi star which is actually c xi. So such that this angle equals this angle note that for both the triangles this and this this angle is common now we have chosen xi star such that this angle angle o xi x is same as angle o x xi star. Therefore by similarity what we get is that the sides are proportional. So what are the sides opposite to this angle here angle x o xi or x o xi star one of them is norm x minus xi other one is norm x minus xi star that will give us this. What is the side opposite to this angle o xi x that is x norm x and what is the side opposite to angle o xi star that is xi star so similarly you get this. So the above equality suggests that norm xi star equal to r square by norm xi. So if you observe norm xi star into norm xi equal to r square how did we choose xi star with some property correct. So xi is here r is in is here xi star is here. So product of the length of xi star and length of xi equal to r square. Hence this xi star is called inversion with respect to the sphere. In real numbers let us look at 0 let us say 1 and I take a point here let us say half what are the inverse of half multiplicative inverse is 2 right. So y 2 into 1 by 2 is equal to 1. So this is exactly the same thing here that is why this is called xi star is the inversion of xi with respect to s of 0 r. Since norm xi star equal to r square by norm xi we get c equal to r square by norm xi square and thus xi star equal to r square by norm xi square into xi. Thus it follows that this equality for all x such that norm x equal to r holds with q equal to 1 and c is such that e power c equal to norm xi by r. Therefore we have phi equal to 1 by 2 pi log norm x minus xi star plus 1 by 2 pi log norm xi by r which if you take 1 by 2 by common it becomes this now it is like log a plus log b that means log ab. So you have this and substituting the value of xi star we get this. So this is phi of xi. So our search for Green s function proceeded by the decomposition formula g equal to k minus phi now we know phi also k is already known therefore g is known. So g has this expression given in this equation. Now let us discuss Poisson's formula for the solution to Dirichlet boundary value problem on the ball and also on the disk in Archer. So recall from lecture 6.2 we made a remark there that this formula which we have proved it gives a representation of the solution u in terms of the Laplacian u on omega that is fine and its boundary values dou n u and u for Dirichlet problem only u is given and dou n u is unknown therefore the formula given above is not useful for computing the solution. But there is an update now we can use the formula and obtain an expression for the solution to Dirichlet boundary value problem by eliminating the term involving the normal derivative of u that is what we are going to do shortly. So Dirichlet boundary condition is u equal to g for x belongs to boundary of omega. So this formula takes this form if Laplacian u equal to f you have this and dou n u unknown u is known g. Now let us apply Green's identity 2 which is given here with u equal to u and v equal to phi of x psi that will give us an equation which involves dou n u. So thus we have these 2 equations if you add these 2 then the term involving dou n u gets cancelled because on the boundary k minus phi is 0 thus we get a formula for u which involves only g dou n u is eliminated. So the solution to Dirichlet boundary value problem is given by this and the special case when u is harmonic this term drops out and we have just this term. In fact we are going to solve harmonic functions such that u equal to g on the boundary therefore this is what we are going to find a new expression for and dou n g is what all needs to be computed. So compute dou n g for d greater than equal to 3 and d equal to 2 separately and substitute in this and see what the formula we get and that formula is called Poisson's formula. Of course if you are given Laplacian u equal to f but not f is not equal to 0 this will give you the solution this will be the representation for the solution. We have to be very careful this will give the solution means we have to prove something but what we have done so far is if there is a solution which is smooth enough then u has this expression that is what we are doing. So we are getting an expression for solution if exists existence is to be proved. Hopefully the Poisson's formula will help us in proving the existence of solutions to boundary value problem Dirichlet boundary value problem. So for omega equal to ball of radius r centre 0 dou n g computation please do this computation by yourself I just put the value of g this is a constant so this came out. So I have to find out the normal derivative of this which is nothing but gradient dot the normal normal to the ball is along the radius therefore x but if you want unit normal x by r because norm x equal to r this will be unit normal outward normal so therefore dou n is nothing but grad x of this quantity dot x by r. Now it is a matter of carrying out the differentiations please do the computations on your own I will skip the details but I will keep the slide for some time so that you can note on. So this is the final expression we get for dou n g here it is simplified very nicely in fact even when d equal to 2 we are going to see that same formula will come for dou n g. So substituting the value of dou n g into this equation will give us this formula which is called Poisson's formula of course we have handled d greater than equal to 3. Now computing dou n g is left an exercise when d equal to 2 and substituting in this formula for dou n g we get this expression look exactly the same formula d equal to 2 that is it. So Poisson's formula has the same form for all d which is u xi equal to r square minus norm xi square by omega dr integral over the sphere r circle if it is two dimensions of g by norm x minus g power d d sigma r you may also write instead of g u itself after all it is a notation for values of u on the boundary. Now let us look at Hornox inequality so let b of 0 r be open ball with center 0 and having radius r in r d d greater than or equal to 2 and u be a harmonic function which is non-negative it takes values which are greater than or equal to 0 that is very important. And let u be continuous on the closed ball of radius r that means u is continuous up to the boundary then for any x in b of 0 r following inequalities hold. So the value of u x and the value of u at the center of this ball are tied by these inequalities or they satisfy these inequalities proof is very simple it follows just from the Poisson's formula look at this Poisson's formula and then triangle inequality whenever y is on the sphere of radius r that means norm y is r right. So norm x minus y by triangle inequality is less than equal to norm x plus norm y but norm y is r therefore we have this here norm x minus y is greater than or equal to norm y minus norm x which is a consequence of triangle inequality therefore you get this. Now use these inequalities in this denominator you get Hornox inequality. So this is what we get one this is the other side we get this. Hornox inequality did not have this expression what was there here is r power d minus 2 into u of 0 similarly here it was r power d minus 2 into u of 0. So proof of Hornox inequality is complete if we knew that this quantity which appeared on the last slide is r power d minus 2 u of 0. The last equation may be rewritten as this yeah I divide both sides with r power d minus 2 so we get this. Now if you look at what is the left hand side it is you are integrating u on the sphere and this is the surface area of the sphere you are dividing with that so this is the average spherical average spherical mean of u over the sphere. Now we are asking whether that is equal to u of 0 this follows from Poisson's formula. So recall Poisson's formula u of x equal to r square minus norm x square by omega d times r integral over the sphere of radius r u of y by norm y minus x power d d sigma. Therefore u of 0 equal to r square by omega d r because norm x is 0 into the integral on the sphere of radius r u y by norm y power d d sigma because x is 0. But when y belongs to this circle norm y is r therefore we get r square by omega d r power 1 plus d into integral of u on the circle. Thus we have this equation is exactly what we wanted to show. So this concludes the proof of Horneck s inequality. Now as a consequence of Horneck s inequality we are going to prove Lieuville's theorem. What is Lieuville's theorem? If you have a harmonic function on rd which is bounded below then it must be constant. It means no non-constant harmonic functions can be bounded below. Now as a corollary we can also get that if your harmonic function is bounded above then also it must be constant. So let m belongs to r be such that u is greater than or equal to m for all x and rd because we are assuming u is a function which is bounded below. Then look at this function u x minus m this will be harmonic and it will be non-negative therefore Horneck s inequality is applicable for this v. Therefore for every r and for every x in b of 0 r we have these inequalities. These are the inequalities coming from Horneck s inequality. So fix an x in rd take r which is positive such that r bigger than r max and pass to the limit as r goes to infinity because when if you want to pass to limit as r goes to infinity whatever x you take there will be a time after which r becomes bigger than r max anyway. So we have this. So in the inequalities above let us pass to the limit as r goes to infinity. What we get is v of 0 less than or equal to vx less than or equal to v of 0. A rough understanding of this is this is r power d minus 1 right d minus 2 into 1. Here it is also the leading term in r power d minus 1 therefore this limit will be 1. Similarly this limit will be 1 as r goes to infinity. So we have this. Now what does this mean vx equal to v0 since x is arbitrary what we have shown is that v is a constant function as a consequence u is a constant function. Let us summarize we have defined Green s function for Dirichlet boundary value problems. We have obtained Poisson's formula which is a representative formula for a solution which is in C2 in omega and continuous up to the boundary of omega when omega is a ball or a disk in rd. It is not clear if Dirichlet boundary value problem admits a solution. It can be shown that Poisson's formula defines such a solution. As a consequence of Poisson's formula we have established Harnock's inequality and Lieuweil's theorem. Thank you.