 Welcome to this lecture. In this lecture, we are going to discuss boundary value problems for Laplace equation. The outline of the lecture is as follows. First we derive what are called Green s identities which are derived from the divergence theorem and then we introduce boundary value problems associated to Laplace equation and study the uniqueness properties of solutions to those boundary value problems. So Laplace equation in d space dimensions, there are only space dimensions here there is unlike the wave equation. So Laplace equation in d dimensions is given by u x1 x1 plus u x2 x2 plus up to u xd xd equal to 0. The operator on the left hand side is called Laplacian u and denoted by capital delta u that is a standard notation for a Laplacian. This is nothing but the trace of the Hessian matrix of the function u. Hessian matrix recall is the matrix of second derivatives of u. So this is the trace of that that means some of the diagonal terms. Any solution to the above equation is called a harmonic function. Laplace equation in 2 dimensions we are going to deal with in some of the lectures which come later on. When d equal to 2 the independent variables x1 x2 are denoted by xy that is a standard practice and we write boldface x sometimes that denotes the tuple x comma y in R2. Thus Laplace equation in 2 independent variables is nothing but u xx plus u yy equal to 0. The non-homogeneous problem where there is a right hand side not 0 function but a general function f. This equation is called Poisson's equation. So in particular if f is equal to 0 get back this equation. So the difference between Poisson equation and Laplace equation is kind of blurred for us we simply call anything as Laplace equation. Actually Laplacian is this operator which is on the left side of both the equations. Here it is a homogeneous equation here is a non-homogeneous equation and the non-homogeneous equation is usually called Poisson equation. So let us discuss what are the green identities and how to derive them. Greens identities play an important role in the analysis of Laplace equation. They are derived from divergence theorem. So what is divergence theorem? It is stated for a domain omega which is bounded and piecewise smooth domain we are going to use in this form. Let psi be a function which is R2 valued function that means psi has 2 component functions psi 1 and psi 2. It is defined on this domain omega. Omega is assumed to be bounded and piecewise smooth domain we will come to that as soon as we see the formula. And we assume that the psi i's have this property that it is c1 of omega bar intersection cf omega bar. It means the derivative is continuous up to the boundary of omega. So omega closure involves omega union boundary of omega. That means derivative of this function psi i is meaningful on the boundary of omega also. And cf omega bar means the functions are meaningful on the boundary of omega. In fact the function psi i's are continuous on omega bar. Then we have this integral omega domain integral. On the right hand side you have a integral on the boundary. So this is divergence of psi dx equal to integral over boundary of omega psi dot n d sigma. What is n? n is a unit outward normal to boundary of omega. And what is d sigma? It is the surface measure which is coming to the boundary of omega from the omega from the usual measure on omega. So in order that the right hand side is meaningful we need to assume that domain is space wise smooth domain. We will not elaborate further on this because you would have learnt this already in course on multivariable calculus. So green signatories they are derived using the divergence theorem on making specific choices for psi in this conclusion of the divergence theorem. So choosing different size will give you different identities. And omega is assumed to be a bounded piecewise smooth domain. n denotes the unit outward normal to boundary of omega. The functions uv appearing in green signatories are assumed to be of this type c2 of omega bar intersection c1 of omega bar. We have already seen this space on the previous slide. This means that the functions second derivatives are also continuous on omega bar. That is the meaning of c2 of omega bar. To make sure that the integrals appearing in them are meaningful. First of all this integral to make sense it is enough that inside the integrand is continuous on omega bar. And that is guaranteed if psi is c1 of omega bar. Later on we are going to see an integral which features second derivatives as the integrand that is why we are assuming c2 of omega bar. It will be very clear from the context. These hypotheses are made so that it is valid for all greens identities. For some of them we do not require all the hypotheses. So now if you apply a divergence theorem with psi equal to grad u what we get is called greens identity 1. So divergence of psi is now divergence of grad u that is Laplace in u. And grad u dot n is precisely the normal derivative of u dou n u. So this is greens identity 1. Now if you apply divergence theorem with this choice of psi we get greens identity 2. So it is integral over omega v Laplace in u minus u Laplace in v of x dx is equal to integral on the boundary v dou n u minus u dou n v d sigma. If you apply divergence theorem with psi equal to v grad u you get greens identity 3 which is integral over omega of grad u dot grad v dx equal to integral over boundary of omega v dou n u d sigma minus integral over omega v Laplace in u dx. In fact this identity we have used many times before which we called it as integration by parts formula. If you notice here the derivative on v is shifting to the derivative on grad u which is giving you Laplace in u. Grad when it goes to the grad u it becomes divergence. So you get v Laplace in u and this is the boundary term. So let us now discuss some of the boundary value problems for Laplace equation. On a bounded domain omega with piecewise smooth boundary boundary of omega it is boundary of omega is denoted by dou omega. We will consider the following three boundary value problems. What are they? Dirichlet boundary value problem. Unknown function is prescribed on boundary of omega here and then we have a Neumann boundary value problem in which normal derivative of the unknown function is prescribed on boundary of omega. Then we have a Robin boundary value problem which sometimes is also called as third boundary value problem. In this a mix of the unknown function and its normal derivative are prescribed. So linear combination of the unknown function and its normal derivative is prescribed on boundary of omega. So let us see what is a Dirichlet boundary value problem. So this is BVP1 boundary value problem 1 Dirichlet problem. Given functions f and g Dirichlet problem consists of solving the boundary value problem. It consists of solving the Poisson's equation Laplacian u equal to f in omega and u equal to g on the boundary of omega. That means the unknown function should agree with the pre-prescribed function g on the boundary of omega and Laplacian u should coincide with the prescribed function f in omega. This is what we want. What is the meaning of a solution to Dirichlet boundary value problem? Here we have to start assuming something on the data. Let f be a continuous function and g also be a continuous function on omega and boundary of omega respectively. A function phi which is c to omega intersection c of omega bar you will actually see why this hypothesis is coming here on phi. It will be very clear once you see the definition. It is said to be a solution to Dirichlet boundary value problem if Laplacian u equal to f in omega and u equal to g on boundary of omega. This is a statement of the Dirichlet boundary value problem. If the function phi is a solution to Laplacian u equal to f it means that for each x in omega Laplacian phi of x should be equal to f of x. So in order that the left hand side makes sense we are assuming phi is c to of omega that is why c to of omega. Now there is a second condition that is a boundary condition. For each x in boundary of omega phi of x should be equal to g of x. If phi is continuous up to boundary that is c of omega bar then values of phi for points x which are in the boundary is meaningful and asking that it is equal to g of x is meaningful. So that is the reason why we have to put these spaces naturally. Now let us look at the second boundary value problem called Neumann problem. So here given functions f and g Neumann problem consists of solving the boundary value problem Laplacian u equal to f in omega and the normal derivative equal to g on boundary of omega. G and f are prescribed they are given to us. What is the meaning of a solution to the Neumann boundary value problem? Assume that f is continuous on omega and g is continuous on boundary of omega. A function phi which is c to of omega intersection c 1 of omega bar is said to be a solution to Neumann boundary value problem if the function phi is a solution to Laplacian u equal to f that is why we are going to assume phi is c to of omega because of this Laplacian phi at every point x in omega should be equal to f of x. Now we have the boundary condition which involves the normal derivative. Normal derivative is nothing but gradient u dot n it is a directional derivative in the direction of the normal. For that reason we need c 1 of omega bar that means derivatives are also continuous up to boundary and hence this is meaningful. So dou n phi of x is meaningful because phi is c 1 of omega bar and hence we can ask that it should be equal to g x that is why the notion of solution to Neumann problem is like this phi should be in this space. Now let us look at the third boundary value problem which is also known as Robin problem. Given functions f g and alpha is a real number Robin problem consists of solving the boundary value problem equation is Poisson's equation in omega boundary condition is a combination of u and the normal derivative of u u plus alpha times dou n u equal to g on boundary of omega. What is the definition of a solution to Robin boundary value problem? Let f be a continuous function on omega and g be a continuous function on the boundary of omega. A function phi c 2 omega intersection c 1 of omega bar function phi in this space c 2 omega intersection c 1 of omega bar this is natural because this is coming from the requirement of Laplacian phi equal to f and this will come because the requirement of the boundary condition. So this function phi is said to be a solution to Robin boundary value problem if the function phi is a solution to Laplacian u equal to f which means for every x in omega Laplacian phi of x equal to f x. This is the reason why we assume phi is c 2 of omega and second condition is the boundary condition for every x on the boundary of omega the equality phi x plus alpha times dou n phi of x equal to g of x holes in order that the left hand side is meaningful we have assumed that phi is c 1 of omega bar a remark as per the definition of Dirichlet boundary value problem the problem is posed on a bounded domain omega. In fact, the other two boundary value problems are also posed on bounded domains by definition the domain omega is bounded means what enclosed by the boundary of omega boundary of omega omega. For this reason Dirichlet boundary value problem is also called an interior Dirichlet problem. Similarly, the other notions we can say interior Neumann problem, interior Robin problem. On the other hand there are boundary value problems which are posed on domains which are complements of bounded domains Laplace equation is to be solved on a domain omega which is the complement of a bounded domain. For example, my omega is here this is my omega it is a complement of this set which is inside and this is my boundary of omega the this curve is my boundary of omega I am drawing the picture in the plane obviously. So, omega is outside this outside region which is here this is boundary of omega. So, we need to solve Laplacian u let us say equal to f outside and here we are prescribing u equal to g for example, on boundary of omega unknown function is prescribed on the boundary of omega. Such boundary value problems are called exterior Dirichlet problems. Similarly exterior Neumann problems exterior Robin problems. Other kinds of boundary value problems are also possible on different parts of the boundary you prescribe different boundary conditions. For example, this is my omega then I consider this part let us call this boundary of omega 1 omega part 1 this is boundary of omega part 2. So, here I can ask u equal to g here I can ask dou n u equal to some function h. I want to solve Laplacian equation in omega in a equal to f in omega and I ask that on this piece u must be equal to g that is the Dirichlet boundary condition on this piece dou n u equal to h that is a Neumann condition. Such boundary value problems are called mixed boundary value problems we will not be studying such problems in this course. Cauchy-Kowalski theorem this is a comment about the initial value problem. Cauchy-Kowalski theorem guarantees that a solution to an analytic Cauchy problem for an elliptic equation in particular Laplace equation exists and the solution is unique locally. This problem is not always well posed we are going to see towards the end of our discussion on Laplace equation an example of a ill-posed Cauchy problem for Laplace equation. Now, the Neumann problem that we have namely Laplacian u equal to f and dou n u equal to g this is in omega this is on boundary of omega. If this problem has a solution f and g are tied up with some relation. So, you cannot have arbitrary f and g for which the Neumann boundary value problem has a solution. In other words if the Neumann boundary value problem has a solution f and g must be compatible with each other in the sense that we are going to soon describe. For a Neumann boundary value problem to admit a solution the data f and g must be compatible that is written in the form of a lemma it does not mean it is a big result it is a simple observation just to remember the observation so that we can recall whenever we want we record them as a lemma or as a theorem in this case it is a lemma. Let f be C of omega bar we have to assume something more so far in the definition of the problem we needed f in C of omega only but here we are asking for C of omega bar because the tools that we are going to use will involve the continuity up to boundary. In fact, we are going to have certain integrals for them to make sense we need this assumption. And g in C of boundary of omega if you belonging to C2 of omega bar is a solution to Neumann boundary value problem on omega then integral over omega of fx equal to integral over boundary of g g is defined on the boundary so integrate on the boundary f is defined in omega so integrate on omega both of them must be equal. So integrate both sides of this equation Laplace in u equal to f on omega what we get is this okay I have exchanged the sides so integral over f integral of f over omega equal to integral of Laplace in u over omega fine. Now applying Green's identity this right hand side integral can be written as this integral over boundary of omega dou in u d sigma the integral on the right hand side thus becomes this okay same but now I know what is dou in u since u solves the Neumann problem dou in u is g and Laplace in u is f therefore we have this equation reducing to an equation where f is here and g is here so thus we have proved the lemma. In particular if f is 0 that is you are looking at the homogeneous Laplace equation Laplace in u equal to 0 that means f is 0 then integral of g over boundary must be 0. Now let us discuss some uniqueness properties of solutions to the three boundary value problems that we have just introduced the Dirichlet boundary value problem has at most one solution remember here we are not saying about existence at all we are just saying it has at most one solution that means if it has a solution then it has exactly one solution that is the correct conclusion from here. And if you solve the Neumann problem then any other solution looks like u plus c where c is a constant in other words difference of any two solutions to Neumann problem is a constant and for the Robin problem we have to assume something on alpha if alpha is greater than or equal to 0 then the Robin problem has at most one solution recall that Robin problem reduces to Dirichlet problem in alpha equal to 0 so proving 3 is enough to prove 1. So what we do is we just prove 3 and then we prove 2 as I have already mentioned Dirichlet boundary value problem is a special case of Robin boundary value problem it is enough to prove the uniqueness result for Robin boundary value problem please note I am calling here uniqueness result by this what I mean is that this problem admits at most one solution. So assume that u1 and u2 are solutions to the Robin problem the uniqueness proofs always proceed like this you take the difference of u1 and u2 and show that that is 0. So therefore you define a w which is u1 minus u2 we want to show that w is 0 observe that w satisfies Laplacian w equal to 0 because Laplacian u1 equal to f Laplacian u2 equal to f therefore Laplacian w is a difference of Laplacian u1 and Laplacian u2 both of them are f therefore Laplacian w is 0. In other words w is a harmonic function in omega and w satisfies the boundary condition w plus alpha dou n w equal to 0 because u1 satisfies the boundary condition with the same g and u2 also satisfy with the same g therefore the difference will satisfy 0 because this is linear in w. So this is the boundary condition satisfied by w. So using the green side t3 with this is the green side t3 we are going to use this with u equal to v equal to w. So left hand side will be integral omega grad w dot grad w that is mod grad w square. Right hand side Laplacian w is 0 so this term is not there so what we have is this term that is w into dou n w d sigma using the boundary condition we get integral over boundary of omega w dou n w equal to this quantity. Please check that. In the last equation the left hand side is non-negative because integrand is non-negative. Right hand side the integral is non-negative alpha is non-negative therefore the product is non-negative but there is a minus sign so it is non-positive. So we have a non-negative quantity equal to a non-positive quantity which is possible if and only if both the sides are 0. That means what grad w equal to 0 on omega dou n w equal to 0 on boundary of omega. On the last slide we have proved that grad w equal to 0 in omega and also the normal derivative of w equal to 0 on the boundary of omega. Since gradient of w equal to 0 in omega w must be a constant function. Now, since the normal derivative is 0 on boundary of omega if we use the boundary condition we get w equal to 0 on the boundary of omega. But what is w? w is continuous up to boundary w belongs to C of omega bar w is a constant and it is 0 on the boundary therefore w must be 0 in omega also. In other words we have Robin problem has at most one solution. Now let us look at Neumann problem. Let u1 u2 be solutions to Neumann problem. Consider the difference u1 minus u2 call it w. Look at the problems satisfied by w. Laplace in w will be equal to 0 and normal derivative w equal to 0 on the boundary of omega. Now using once again the gradient identity 3 with u equal to v equal to w. We have integral over omega mod grad w square that is coming from the left hand side on the right hand side as before Laplace in w is 0. So, second term drops out what you have is this term which I have written here w dou n w d sigma. But dou n w is 0 therefore this integral is 0. This means grad w equal to 0 in omega that is all we have information nothing more. If grad w is 0 w must be a constant function that means u1 minus u2 is a constant function therefore u1 equal to u2 plus constant. Let us summarize what we did in this lecture. We have derived Green s identities using divergence theorem. Introduce 3 boundary value problems for Laplace equation. Discouraged if there can be more than one solution for each of the 3 VVPs. Thank you.