 Welcome to tutorial 2 on Laplace equation. Problem 1, let D of 0, root 2 denote the disc of radius root 2, open disc of radius root 2 and S of 0 root 2 denotes the circle of radius root 2. Consider the Dirichlet boundary value problem Laplace nu equal to 0 in the disc D 0 root 2 and U of xy equal to 1 plus 3 sin 2 theta for xy belongs to S of 0 comma root 2. Find U 0 0, if you look at this problem it looks somewhat strange because you are prescribing U of xy and you are not giving a function of xy here you are giving a function of theta. So usually this is how people may post sometimes but then you have to realize that because the disc theta is that angle which is coming into the picture and in the background there is r cos theta r sin theta kind of description for the circle x square plus y square equal to r square. So now how do you find this U of 0 0? One way of course is to find out the solution of this problem and find the value at 0 0 but do we have to do that much? Idea is use mean value property U is known on the circle you are asking what is the value of U at the center and precisely the mean value property is what connects both the things. So U of 0 comma 0 equal to 1 by 2 pi r here r is root 2 S of 0 root 2 1 plus 3 sin 2 theta into d sigma of xy. We will expand this integral so that is equal to 1 by 2 pi root 2 integral from 0 to 2 pi because the angle theta runs from 0 to 2 pi and root 2 d theta therefore that is nothing but 1 by 2 pi 0 to 2 pi 1 plus 3 sin 2 theta d theta that is equal to 1 by 2 pi the integral can be split into 2 integrals namely 0 to 2 pi of d theta plus 0 to 2 pi of 3 sin 2 theta d theta this is 0 therefore we get 1 by 2 pi into 2 pi that is equal to 1. Let us move on to problem 2 let omega be the complement of the closed unit disc centered at origin that is set of all xy and r2 such that x square plus y square is greater than or equal to 1. Let U be C2 omega intersection C of omega bar function such that Laplacian U is 0 that is U is harmonic in omega and such that limit of U as norm xy goes to infinity is 0. Show that the maximum of modulus U xy as xy varies in omega bar is same as maximum of mod U xy as xy varies in boundary of omega. In other words we are saying the maximum of mod U over omega bar is attained on the boundary rather is definitely attained on the boundary of omega. The requirement here suggests application of maximum principle namely the weak maximum principle but we do not have that on unbounded domains and our omega is unbounded domain let us see how to do that. Let us see how to cleverly apply the weak maximum principle and then show what is required in this problem. So what is the idea to solve this problem? We are given that limit of U is 0 as xy goes to infinity that is this is 0. So that means that one can make mod U as small as we want outside a big enough disc that is for example this and you can make mod U small here. So that will tell us that the supremum of U is meaningful because U is now bounded and then that actually becomes maximum of U inside this region and that is when we plan to apply maximum principles and get the answer. Of course recall this picture is not exactly correct because we are our domain is complement of the unit disc now it is fine. So the natural choice would then to consider this annular region and apply some kind of maximum principles. So let us denote this circle by D0r this disc and this disc anyway is D01. So look at this annual region D0r minus D01 if you call this omega r. So we can apply a maximum principles and conclude that the maximum of mod U it can be minimum of U or maximum of U that is attained on the boundary and we conclude that it cannot be this boundary because we have made it very small. The r is chosen such that it is small very small here therefore it has to be on this and that is what exactly what we want to show. We will go into the details now. So since U belongs to C of omega bar and the limit is given to be 0 we have look at modulus of U of x, y as x, y vary in omega of course this is a subset of r this set is bounded set of course non-empty therefore it has a supremum. So let that m denote the supremum now suppose m is 0 then there is nothing to prove if m is 0 it means U itself is 0 therefore what we are supposed to prove holds automatically that nothing to prove. So we would as well assume that m is a positive quantity and show what is required namely that maximum of mod U on omega closure is same as maximum of mod U on boundary of omega boundary of omega recall is the unit circle. Now in this case since the limit is 0 limit of U is 0 there exists r positive such that modulus of U of x, y we said it can be made as small as I please so I will make it less than m by 2 and this will hold for every x, y with what property x, y is outside the disc of radius r that is for every x, y such that norm x, y is bigger than or equal to r. So now we are going to take that r which we have just chosen here by this constraint and then we propose the annular region omega r denote this by disc of radius r minus the disc with centre 0 radius 1. r satisfies r has a following property that whenever norm x, y is bigger than or equal to r mod U x, y is less than m by 2 this is the way we have chosen r. Now let me recall the notation again omega r denote this by disc of radius r minus the disc with centre 0 radius 1. So this is an annular region so this is disc of radius 1 is disc of radius r and this is my domain omega r. So what is the properties now I have on this domain omega r the value m is attained by mod U on omega r and Laplacian U equal to 0 in this annular region because it is given to be 0 on omega this is a subset of omega therefore Laplacian U 0 and U is C2 of omega r intersection C of omega r closure this is also true. Now what is this m? m is maximum of mod U over omega r bar also therefore m is either maximum of U on omega r bar or maximum of minus U on omega r bar. Now by strong maximum principle of course we have stated only for strong maximum principle we can also state a strong minimum principle as a consequence we can deduce that from strong maximum principle. So that tells us that whether it is minimum or maximum it is attained inside the domain it has to be a constant function the function U that we are dealing with is not a constant function we know that because the behaviour of U at infinity we assume that limit of U is 0 at infinity. So if U had been a constant function then the function should have been the 0 function that is the situation when m equal to 0. Now we are in the case where m is positive therefore U is not a constant function therefore the maximum of U or minimum of U is attained only on the boundary of omega r by strong maximum principle and strong minimum principle. So m is attained either on the circle of radius r or on the circle of radius 1 that is this or this but what happens on the circle of radius r? Circle of radius r mod U is strictly less than m by 2 therefore m cannot be attained on the circle of radius r therefore m is attained only on the circle of radius 1 that is maximum of mod U or omega bar is equal to maximum of U mod U on boundary of omega. Problem 3 let omega be a bounded domain let U belongs to c to omega intersection c of omega bar be such that Laplace in U equal to minus e power U in omega and U equal to 0 on boundary of omega show that U of x is greater than or equal to 0 for all x in omega. So how to solve this problem we know nothing about this kind of equations right this is actually a non-linear equation or if you want to be milder same linear equation but then we have no idea how to handle this. So we start thinking the other way suppose this is not true U of x is greater than or equal to 0 is not true it means at some point in omega U is less than 0 in particular it means the minimum is going to be a negative number but here if you see U is 0 on the boundary therefore a minimum of U on omega closure is attained at a point in omega at points of minimum what is the Laplace in U has to be greater than or equal to 0 but by this equation Laplace in U is always less than 0 therefore it cannot happen that U is less than 0 at some point in other words U is always greater than or equal to 0 we are going to write down the details as such this problem is not something to do with Laplace equation but the fact that Laplace in U has a certain sign at points of minima or maxima since at every point of omega we have Laplace in U which is equal to minus e power U and that is strictly less than 0 you cannot minimum of U on omega bar inside omega the minimum cannot be achieved inside omega that means minimum achieved only on the boundary of omega but on the boundary of omega U is 0 so minimum of U on omega bar is same as minimum of U on boundary of omega but that is 0 but saying that minimum of U on omega bar is 0 is same as saying that U is greater than or equal to 0 for all x in omega so this is a problem featured in Laplacian but the solution needs knowledge of only maxima minima in multivariable calculus. So problem 4 let P of xy be a polynomial of degree k let U belongs to C2 of R2 be such that Laplace in U is 0 that is U is harmonic in R2 and such that U of xy equal to P of xy for all xy on the unit circle that means we have a harmonic function in R2 which equals a prescribed polynomial P on the unit circle then show that U itself is a polynomial and polynomial solutions of Laplacian equal to 0 are called harmonic polynomials. We are going to discuss two approaches to solve this problem the first approach is not successful but nevertheless we mention the ideas we think of using mean value property and also the fact that harmonic functions are known to be real analytic functions okay how do we plan to use this idea or why do we think of this idea we are given that U equal to P on the unit circle P is a polynomial if P is a polynomial of degree k you know k plus 1th order derivatives of P are 0. So therefore we if you think of d alpha U at the origin 0 0 for mod alpha bigger than or equal to k plus 1 by mean value property they will be 0 because mean value property is written in terms of an integral on the circle but on the circle you have d alpha P whenever alpha is such that mod alpha is bigger than equal to k plus 1 d alpha P will be 0 because P is a polynomial of degree k therefore we get that d alpha U at 0 0 will be 0 whenever mod alpha is bigger than or equal to k plus 1 and since U is supposed to be real analytic function I can write the Taylor series expansion at the point 0 0 of course this will be a finite expansion in other words it is a polynomial it is not a series it is going to be just polynomial and then I hope to show that that is a polynomial which we are looking for let us see what are the problems in this approach. So let me just write what is the idea mean value property and real analyticity. So mean value property gives us that d alpha U at 0 0 equal to 1 by 2 pi integral on the unit circle d alpha U at x y d omega of x y that is nothing but 1 by 2 pi s of 0 1 d alpha but on the unit circle U equal to P therefore it is d alpha P of x y d omega of x 1 this is true for all alpha that is mod alpha greater than or equal to 0. But since P is a polynomial of degree k whenever mod alpha is bigger than or equal to k plus 1 we get d alpha U at 0 0 is 0 because d alpha P is 0. So writing down the Taylor series what we get is U of 0 0 I will write only a few terms dou by dou x at the point 0 0 into x dou by dou y at the point 0 0 into y plus second order derivatives dou x dou y this will be 2 at 0 0 x y plus dou 2 U by dou y square square into 1 by 2 and so on. So if you want to understand the Taylor series you can consult any book on multivariable calculus in particular the book of Bartle or Rudin. Rudin's book on mathematical analysis and Bartle's book on elements of real analysis we will find that. Of course this will be valid in a neighborhood of 0 0 or in a disk containing 0 0 but as we observed this is going to end after sometime. It is not a series it is going to be a polynomial so there is no problem about valid it will be valid everywhere. So we cut hold of a polynomial expression for U of x y we have to show that this U is indeed Laplacian U equal to 0 and U equal to P on the unit circle. We are not applying any uniqueness theorems if we had uniqueness to this problem the given problem we can start to say that the U we got is answer so the trouble lies there I am going to explain this later. So let us call the polynomial that we got in the previous slide as Q of x y using the real analyticity of the function U. So now Q satisfies Laplacian Q equal to 0 in the unit power and Q of x y equal to P x y for all x y such that x square plus y square equal to 1. The question is why this is what we would like to say but is this true one seems to be okay because one means this this seems to be okay because Q is actually a representation of the U and we know Laplacian U equal to 0 therefore Laplacian Q is 0. But why is this true Q x y equal to P x y no idea it looks like the proof is going to break down here. So we go for another approach approach 2. Let us look at a simple case suppose that the given P of x y okay is such that Laplacian P equal to 0 that means P itself is a harmonic polynomial then what happens U of x y is equal to P of x y the unit disk because Laplacian of U minus P equal to 0 in the unit disk and U minus P equal to 0 on the circle therefore by strong maximum principle we get U minus P equal to identically equal to 0. So in other words U is identically equal to P in particular U is a polynomial done. So trouble is that this P which is given may not be harmonic polynomial then what you do in that case. Just notice some interesting thing here which is if U x y equal to P x y on the unit circle U x y equal to P x y plus 1 minus x square minus y square into L of x y for any polynomial L that is because 1 minus x square minus y square is 0 on S01. So this will happen on S01. Therefore we ask can we supply something from here into this so that we get a harmonic polynomial here and something here let me write a tilde for the moment. So in other words we want to write whether P of x y can be expressed like this where H is harmonic and you have addition of this which is vanishing on S01. So suppose we are able to write P of x y equal to H of x y plus 1 minus x square minus y square into L of x y where H and L are polynomials course polynomials and Laplace in H is 0 that is H is a harmonic polynomial in B01 let us put for a polynomial it does not matter once Laplace in H is 0 Laplace in H is 0 everywhere U of x y equal to P of x y on S01. Therefore by strong maximum principle we get U identical equal to H. So this follows from the observation that we made on the previous slide when the given P were harmonic polynomial then we concluded that U is identical equal to P. Now here I will treat this as the given polynomial H and U equal to H on the this is actually equal to H right U equal to H on the unit circle then H is the solution by strong maximum principle. So U in particular is a polynomial in B01. So the question now is how do I catch hold of H and L as on the last slide that is P equal to H plus 1 minus x square minus y square into L that H and L how do I get of course if you get L you have the H. So let degree of P in fact it is given to us equal to K. So it is enough to find L such that Laplace in of P is equal to Laplace in of 1 minus x square minus y square into L of x y and degree of L is K minus 2 because I want to write P is equal to H plus 1 minus x square minus y square into L. So this is already a degree to polynomial so L is degree K minus 2 because P is given to be degree K. So this is an actual condition we get because I want Laplace in H equal to 0 that is if and only if Laplace in of P is equal to Laplace in of this quantity that is what I have written here. Thus we are interested in solving this equation here notice the right hand side is known P is given polynomial so this is known. So we want to solve this now it looks like equally difficult problem. Luckily it is not that difficult because we are only looking for L which is polynomial of degree less than or equal to K minus 2 that really helps us because if you are looking at only polynomials degree less than or equal to K minus 2 it forms a vector space of finite dimension that makes our job easy as we are going to see on the next slide. So define T it is a mapping from polynomials of degree less than or equal to K minus 2 to same space polynomials of degree less than or equal to K minus 2. What is operator? It takes a polynomial H of x, y to this multiply with this 1 minus x square minus y square then you get a polynomial of degree less than or equal to K and then take Laplacian of that so that will be a polynomial. So this operator is well defined it is a linear operator and this is a finite dimensional vector space therefore what we know is T is 1 to 1 if and only if T is on 2. Now what we are interested in is the T is on 2 because if T is on 2 I can find an H such that this quantity is equal to Laplacian P because P is a degree K Laplacian P will be of degree less than or equal to K minus 2 that is how I get this H which I will call it as L. Once I have L I have my capital H therefore we will show that T is 1 to 1 it is easier to show T is 1 to 1. So suppose that Laplacian of 1 minus x square minus y square into H of x y is 0 in B01. In other words what I am going to show that the kernel of T consists of only the 0 element. So I assume that H belongs to the kernel of T that means this equation is satisfied everywhere but I have written in B01 because I have plans to apply some maximum principles. H has this property that means this is the function I am looking at let me call it as V. So V is a harmonic function in 0 1 B01 and what is its value on S01 it is 0 on S01 because on S01 x square plus y square equal to 1 therefore V is 0. Now the maximum principle tells me that V has to be identically equal to 0 that means 1 minus x square minus y square into H x y is the 0 polynomial or 0 function or the 0 polynomial that implies that H of x y is 0 polynomial. So we have shown that T is 1 to 1 therefore T is 1 1 and that implies T is on 2 therefore there exists L of x y polynomial of degree less than or equal to k minus 2 such that Laplacian of 1 minus x square minus y square into L of x y minus P of x y is 0. So define or call H of x y equal to this P of x y minus 1 minus x square minus y square into L x y. Now what can we say about H? H is a polynomial Laplacian H is 0 of course in particular in B01 and H of x y equal to P of x y on S of 0 1 that is for x y such that x square plus y square equal to 1 therefore by strong maximum principle or the uniqueness or uniqueness of solutions to Dirichlet border value problem we get that U of x y equal to H of x y on B01. A natural question because we are dealing with polynomials is U of x y equal to H of x y for all x y in R2 we have only shown the equality on B01 because we applied strong maximum principle on the domain D01 on the disc of radius 1. Answer is S because U equal to H on the disc of radius 1 center origin already this we already showed we will not prove this. A more general fact is that if U and V are real analytic on R2 and distinct then look at the set x in R2 such that U of x equal to V of x this set has measure 0 is more general. So we already have seen that this set in our on our example or in our problem this set contains the disc of radius 1 already which is non-zero measure therefore we can apply this and also say that U equal to H in R2 itself.