 Welcome, in this lecture we are going to look at a few more consequences of mean value property. Outline of lecture is as follows, first we prove that uniform limit of harmonic functions is harmonic. This is sounding like uniform limit of continuous functions is continuous exactly like that uniform limit of harmonic functions is harmonic and then harmonic functions are analytic. Then we are going to look at two imposed problems for Laplace equation. We will see one problem without solutions and we see one problem where there is existence uniqueness but the third requirement of Hadamard phase and thereby we get imposeness. It is an example of Hadamard. Then we show that Dirichlet boundary value problem is well posed for Laplace equation. So, uniform limit of harmonic functions is harmonic. It is also called Harnack's theorem. There are too many theorems going after this name Harnack. So, you have to be careful when referring to these theorems. So, this theorem says take a bounded domain omega in R2 and take a sequence of functions which are C2 of omega and C of omega bar which are harmonic on omega. That means Laplace and Un is 0 for every n and to make sense of that we need C2 and C of omega bar it means we are going to talk about the values of Un on the boundary of omega. So, that is going to come now such that Un equal to fn on boundary of omega which means fn's are prescribed and then Un's are solutions to Laplace and Un equal to 0 satisfying this Dirichlet boundary condition. Suppose fn goes to f in C of boundary of omega it means modulus of fn minus f goes to 0 uniformly as x varies in boundary of omega. So, it is a uniform convergence say omega is a bounded domain boundary of omega is a compact set continuous functions and compact set you can define maximum norm of such functions supremum norm it is called sometimes it is actually the maximum. So, maximum of mod fn minus f as x varies in boundary of omega goes to 0 that is the meaning of uniform convergence. Then the sequence Un converges uniformly on omega closure that means Un also converges uniformly if the boundary values converges uniformly to a function U that means there is a such a function U to which Un converges uniformly and U is harmonic that means Laplace Un is 0. If you look at uniform convergence if you have a sequence of whatever smoothness you have if it converges uniformly you can expect the limit to be only continuous. But now we are saying U is harmonic that means Laplace Un equal to 0 that means something needs to be proved that U is twice differentiable so that Laplace Un makes sense and then further show that Laplace Un equal to 0 and U equal to f on the boundary of omega. Let us turn to the proof of the theorem fix m and n then Un minus Um is harmonic on omega because it is a difference of two harmonic functions and belongs to C of omega bar because both Un and Um belongs to C of omega bar. On the boundary of omega Un minus Um is nothing but fn minus fn because Un is fn on the boundary Um is fm on the boundary. Since fn converges uniformly on boundary of omega it is a Cauchy sequence in C of boundary omega space that is the uniform metric. Uniform metric means distance between two functions let us say f and g in C of boundary of omega is defined as maximum as x varies in boundary of omega of modulus f of x minus g of x that is what is called the uniform metric. The idea is to show that Un is a Cauchy sequence in C of omega bar and stability estimate connects Un with fn. Applying the stability estimate which we have proved in lecture 6.4 we get this. This is the stability estimate because Um minus Un is fm minus fn on the boundary so maximum of Um minus Un on omega closure is less than or equal to maximum of Um minus Un which is fm minus fn on the boundary of omega. Now we know that this is a Cauchy sequence therefore this will be a Cauchy sequence. Since fn is Cauchy in C of boundary of omega for a given epsilon you can find a capital N such that this can be made as less than epsilon whenever m and n are bigger than or equal to this n. This is a definition of a Cauchy sequence. In view of the inequality that we have written down on the last slide which is coming from the stability estimate we have this. Now in view of this this will tell us that maximum omega closure mod Um minus Un is also less than or equal to epsilon in fact less than epsilon for every m and n bigger than or equal to n. So therefore Un is a Cauchy sequence in C of omega bar and hence converges uniformly. Why a Cauchy sequence converges because this space is a complete space it is a complete metric space or it is a Banach space. Therefore every Cauchy sequence converges in particular Un converges. Call the limit as U. So U will be an element in this space C of omega bar. So let U belongs to C of omega bar be the limit of Un. But now we want to show that U is a harmonic function that means we have to somehow show that U has two derivatives to start with. Since each member of the sequence Un has mean value property because Un are harmonic functions. The function U will also have a mean value property reason the convergence Un going to U is uniform and uniform convergence tells us that we can swap integrals and limits. Thus U is a continuous function and has the mean value property. Therefore U is harmonic we have shown this earlier a continuous function which has mean value property is harmonic function. We did this in lecture 6.7 and U equal to f on the boundary of omega follows from Un equal to fn on boundary of omega. Now let us show harmonic functions are analytic. Let U be a harmonic function defined on a domain omega in R2. Then U is analytic what is the meaning of analytic? Given any point in omega there is a disc around that point on which U has a Taylor series representation. Outline of the proof of the theorem we are not going to prove the theorem itself. So we proved in lecture 6.7 harmonic functions are c infinity. Let x naught belongs to omega be fixed. If U is analytic the Taylor series for U about the point x naught should converge in a disc around x naught that is the definition. Since U is c infinity of omega Taylor series can be written down. So this is a formal Taylor series we need to show that this series actually converges to Ux at every point x in a disc around the point x 0. Since the function is c infinity this can be written down because each other term can be written down. Essentially what we know is d alpha Ux naught is meaningful because U is c infinity. Proof of convergence of the above series follows from suitable estimates on the derivatives because these are very generic term they are not going to help you much. This is what is required d alpha U it should have some decay estimates. We skip the proof as it is very technical thus harmonic functions are analytic. Initial value problems without solutions. So example of an ill posed problem we are going to see. Example 1 let gamma denote a segment of x axis. Let G be a function defined on gamma. Then this Cauchy problem Laplacian U equal to 0 in the upper half plane x y in R2 so is that y positive. So this is the upper half plane. And on the x axis we have given the Cauchy conditions U of x 0 is 0 and dou U by dou y at x comma 0 is Gx it is given for all x 0 belonging to gamma that is points of gamma. It has no solutions unless this function G of x itself is an analytic function. Sometimes we call it real analytic function just to distinguish it from the analytic functions of the complex analysis. Let P belongs to gamma. Let D be an open disk with center at P such that x axis cuts D into two equal parts. It means that disc is symmetric about the x axis and D intersection x axis is contained in gamma. Recall gamma is subset of x axis on which the function G is defined. Let D plus denote the part of D which lies above x axis. That means set of all x y in D such that y is greater than or equal to 0. So let U be a C2 function in D plus closure which is a solution to the Cauchy problem. Let U be extended to the whole of D as U of x y equal to minus U of x of minus y per x y belongs to D and y negative. Points x y belonging to D have to have y positive or negative. If y is greater than or equal to 0 U is already defined. So for y less than 0 we will define. What we do is that if y is less than 0 take minus y that is positive therefore U of x minus y meaningful and put a minus sign in the front. So this is the definition of U. The extended function has the following properties. U is a C2 function of D and Laplace in U equal to 0 on D. Because of these two requirements we have put a minus sign here. If you do not put minus sign you will not get this. Checking is very simple and is left as an exercise. So proved in lecture 6.7 harmonic functions are analytic. In particular U is analytic at P and so is its derivative dou U by dou y. Thus G which is the restriction of an analytic function to x axis is itself analytic or real analytic. Now let us look at the Hadamard example on ill poseness of initial value problems for Laplace equation. We proved that Cauchy problem for wave equation is well posed and domains like this x belongs to R and t belongs to 0 t and this happens for every t positive. Let us consider a similar problem for the Laplace equation now. Consider the following Cauchy problem posed in the upper half plane. U xx plus U yy equal to 0 for xy belonging to R cross 0 infinity that is x belongs to R y positive. U of x0 equal to fx for all x in R. U y of x0 equal to gx for all x in R. This is the Cauchy data f and g are the Cauchy data. Note the initial conditions are prescribed in exactly the same way as it was done for wave equation. In case of wave equation the Laplace equation here is replaced with wave equation. The Cauchy conditions remain the same. For the moment let us agree that solution to Cauchy problem exists and is unique. In fact we are going to consider a specific f and g where we explicitly know the solutions but for discussion sake let us assume that the solution exists and is unique. This assumption means that if we somehow find a solution of the Cauchy problem then that is a only solution. This also means that Cauchy problem is ill-posed because the third requirement of Hadamard is not met. If the problem were to be well posed the following stability estimate is expected to be satisfied. This is exactly the same way we have written for the wave equation I am writing here. So given epsilon positive there is a delta positive such that whenever the data are delta close in some uniform sense then solutions remain epsilon close in some uniform sense. What should we do to prove that solutions to Cauchy problem do not satisfy stability estimate? Formulate the negation of the stability estimate which is there in the previous slide and convince yourself that we are proving an equivalent statement. We are going to produce two sequences fn and gn the Cauchy data is f and g. So we are going to take in place of f and g fn and gn respectively. So we are going to produce a sequence of Cauchy data f and gn which are close to the zero function but the corresponding solutions are far from the solution u identically equal to zero. When Cauchy data is zero the Cauchy problem for the Laplace equation has zero as a solution. So this is that solution which is the solution with zero Cauchy data. Let fn gn be given by fn is identically equal to zero gn is sin nx by n for x in R. Why are we taking fn equal to zero because if something fails it fails magnificently therefore we consider therefore we consider only gn. Note that they are very close to the function zero it is very obvious one is anyway zero this uniformly goes to zero mod gnx is mod sin nx by n which is less than or equal to one by n that goes to zero. So gn for large n are very very close to the function zero. The solution to Cauchy problem with the above Cauchy data is given by this formula one can easily check there is a solution to the Cauchy problem for the Laplace equation. Now for n large the initial conditions are very close to zero we have already observed and hence can be thought of as a perturbation of the zero initial state. However the sequence of corresponding solutions unxy given here is not uniformly bounded on the domain r cross 0 t whatever may be the t that you fix. The reason is the presence of the hyperbolic sin term here sin hyperbolic function in the expression for unxy the stability estimate fails. A remark on example 2 this example shows the difference between the Cauchy problems for wave equation and Laplace equations posed on the upper half plane. Even though the nature of the Cauchy data imposed is the same changing the equation from wave to Laplace changes the stability property drastically. Directly problem is well posed for Laplace equation. We have proved in lecture 6.4 it was called stability estimate this theorem. So let omega inside r2 be a bounded domain for i equal to 1 to 2 let ui belongs to c2 omega intersection c of omega bar solve the Dirichlet boundary value problem given here Laplace in ui is equal to f on omega ui equal to gi on boundary of omega. Then we have proved this stability estimate. This result shows the third requirement of Hadamard is met by the Dirichlet BVP. Because if this data is closed solutions are closed. This is exactly the continuous dependence on the boundary data for the Dirichlet boundary value problem. Let us summarize what is done in this lecture. We proved that uniform limits of harmonic functions are also harmonic presented 2 examples of we will pose problems for Laplace equation. Both of them are like initial value problems or Cauchy problems presented a well posed problem for Laplace equation. Thank you.