 Welcome. In this lecture we are going to discuss fundamental solutions in RD for the Laplacian. The outline of the lecture is as follows. First we introduce the idea of a fundamental solution. Then we move on to find fundamental solutions for Laplace operator in RD and then we study some properties of fundamental solutions. Fundamental solution what is it? Why is it fundamental? So what is a fundamental solution? Let us start with a matrix analogy. Let A be a d by d invertible matrix and B be a vector in RD. Consider the linear system Ax equal to b. b is given you want to find solution for x. Since A is invertible it has exactly one solution we know that. Imagine you have a factory which sells solutions to the linear system Ax equal to b. Whenever a customer gives you a specific b you will give him x to the customer. Will you solve every time a customer approaches you with these b? That means whenever a customer comes and gives you b you try to go and find solution for x by your own method how to solve the system and as a customer varies you have to solve the system again and again. Will you do that or do you have a smart way of running your factory? So what is a fundamental solution? Find solutions corresponding to a few selected b for the linear system Ax equal to b. Solve the system for b in E1, E2 up to Ed that means for b equal to E1 you solve, b equal to E2 you solve Ax equal to E2 similarly you solve up to Ax equal to Ed that means the d times you solve this system. What is even E2 Ed? It is the basis for RD. Let us say we take the standard basis even E2 Ed for RD. So E1 will be the d tuple where the first component is 1 rest of them are 0, E2 is the second component is 1 rest of them 0 and similarly Ed is the d th component is 1 and first d minus 1 components are 0. We know that this is the standard order basis for RD. So for each of the basis elements you solve Ax equal to b. Express any other b which the customer gives you as a linear combination of these basis vectors which is very easy. If b is equal to b1, b2, bd then b is nothing but b1 E1 plus b2 E2 up to bd Ed. If b is a vector like this then b is nothing but b1 E1 plus b2 E2. So you know the readily what are the coefficients which are appearing in this combination. For Ax equal to b is a linear combination of these solutions xis which are solving Axi equal to Ei. So the set x1, x2, xd may be called a fundamental set of solutions in the context of Ax equal to b for obvious reasons. We are on the lookout for a collection of functions associated to the Laplace operator which mimic this set x1, x2, xd in the case of Ax equal to b. So why is fundamental solution so named? We will get a fundamental set of solutions for Laplace operator having an infinite number of functions. So the set we are going to get for Laplace operator will consist of infinite number of elements unlike the case of linear system Ax equal to b where it had only d number of elements. It is not a surprise as function spaces are infinite dimensional unlike rd which is finite dimensional. Any solution to Laplace in u equal to f is expected to be a superposition of the solutions from the fundamental set. Some in rd will be replaced by an integral. We are going to see this. The word fundamental set is often used as a substitute for a basis. So fundamental solution for Laplace operator in rd. Laplace equation is invariant under any real orthogonal transformation. What does that mean? Let m be a d by d orthogonal matrix that is m transpose m equal to identity matrix. Define a change of coordinates on rd using this orthogonal matrix m by this set y equal to mx. x is your original coordinate system you are introducing new coordinates y, y equal to mx. Let u be denoted by u of x. Define a function v a function of y by this v of y equal to u of m transpose x. Let delta x and delta y denote Laplacian in the x coordinate system and y coordinate system respectively. Conclude. So this is going to be an exercise. We have done NF exercises on change of variables and how it affects a PDE, how the PDE gets transformed under a change of variables. Conclude the invariance of Laplacian under orthogonal transformations. That is Laplacian with respect to x variables of u is same as Laplacian with respect to y variable of the function v where y and x are related by y equal to mx. In particular Laplacian is invariant under rotations. Thus it is natural to look for solutions to Laplacian u equal to 0 which have rotational symmetry. Whenever we are looking for solutions on domains which themselves have this rotational symmetry that is rotationally invariant. For example, rd trivially balls in rd and annular regions in rd. So finding fundamental solution in rd. How do we do that? Fix a point xi in rd. Look for solutions to Laplacian u equal to 0. Having this form that v xi of x because xi is fixed. So for every fixed xi in rd we are going to find solution. v xi of x equal to xi of r. See this already suggests we are going to find as many functions as the elements in rd. So look for solutions to Laplacian u equal to 0 having this form v xi x equal to xi of r. What is r? r is nothing but norm x minus xi that is the distance from x to the fixed point xi which is given by this formula of course. This is Euclidean norm. Therefore, this is equal to square root of i equal to 1 to d xi minus xi square. Substituting the formula for v xi in Laplacian u equal to 0 yields Laplacian v xi of x equal to psi double dash of r plus d minus 1 by r into psi prime of r and that is equal to 0. This is what we want. Therefore, finding v xi is same as finding psi and psi satisfies this ODE. So we need to solve this ODE. This is a second order ODE with variable coefficient, but it is a simple variable coefficient so it is very easy to solve. So from this equation psi double dash of r plus d minus 1 by r into psi dash of r equal to 0 which was obtained on the last slide. We get psi dash of r equal to constant times r power 1 minus d because there is no term psi in this equation without derivative. You set psi dash of r equal to some g of r then this will be a first order ODE. You can solve that and you get this expression. So therefore, psi dash of r equal to constant times r power 1 minus d. Integrating the last equation, we get psi of r equal to c times log r if d equal to 2 and c by 2 minus d r power 2 minus d if d is greater than or equal to 3. So therefore, the form looks different in dimension 2 and dimension bigger than or equal to 3. This is the reason why we will be considering d equal to 2 separately and d greater than or equal to 3 separately in our analysis in the next 2 lectures. So in terms of x coordinates v psi of x is c times log r is norm x minus zine. So substituting r equal to norm x minus z we get this expression v psi of x. So now we are ready to define what is called fundamental solution for Laplace n or fundamental solution for Laplace operator in Rd. The fundamental solution for Laplace n is this function k it is a mapping from rd cross rd minus diagonal you are removing a set from rd cross rd a certain set which we will define to r defined by exactly the same formula as before. So we have to simply mention what is a diagonal. Diagonal stands for all those x comma xi in rd cross rd such that x equal to xi. So remark on the function k of xi for each fixed xi in rd the function x going to k of xi satisfies Laplacian k of xi equal to 0 for every x different from xi. When x equal to xi there is a problem it is not defined k is not defined but for any other x Laplacian k of x comma xi equal to 0. Thus k is a solution to Laplace equation on rd except for this xi. The family of these special solutions that is the family is indexed by xi in rd this family generates all solutions to a Laplacian u equal to f that is why k is called the fundamental solution. Now compare the analogy that we have given in the case of system of linear equations fundamental set there were finitely many there x1, x2, xd here we have this family of functions indexed by xi in rd. We state this result and we do not prove the result. Let us look at some properties of fundamental solutions that is a theorem. Let k of xi denote the fundamental solution for Laplacian we have already defined this on an earlier slide. Let omega be a smooth bounded domain in rd let xi belongs to omega for u belonging to c2 of omega bar the following identity holds that u of xi is equal to integral over omega of kx xi Laplacian u dx minus integral over boundary of omega k dou nu minus u dou nk d sigma. If u is c2 of omega bar and harmonic in omega that means Laplacian u equal to 0 then the first term will drop out then you have only this term. Then for xi in omega we get u xi equal to this integral which is the second term here. So once you show one two follows immediately and the following quality holds in the sense of distributions on rd that is Laplacian k of xi equal to delta xi. What we saw is Laplacian k of xi equal to 0 whenever x is not equal to xi. Now there is always this question what is that what happens at x equal to xi. So this is that is the effect here delta xi comes in delta xi is the derived delta in case you do not know this you can ignore I am going to explain what this means. This means that for every phi in c0 infinity of rd the following equality holds. So loosely speaking multiply with phi and integrate integral phi delta will give you phi of xi and here you do integration by parts transfer the Laplacian from k to phi and you get this. So phi xi equal to integral over rd of k of xi delta phi of x dx proof of one let u belongs to c2 of omega bar and xi be a point of omega. Note that we cannot apply Green's identity to directly with v equal to kxz we would like to do that but we cannot do that why because k is singular at x equal to xi there is trouble for k at x equal to xi and here if you are trying to use v equal to k you have Laplacian k that will not be integrable. So there will be such problems so we will not do that what we will do is we somehow remove this point xi so we cut out a ball b of xi rho from omega then everything is alright along with its boundary cutting a ball along with this boundary means cutting this closed ball. Recall this is the notation we were using b closed xi comma rho means it is all those points which are at a distance less than or equal to rho from the point xi here it is strictly less than for the open ball this is a closed ball and then we will apply Green's identity 2. So like omega rho b omega minus the closed ball Green's identity 2 with v equal to k of xi on the domain omega rho reads as this is exactly Green's identity 2 I have just put v equal to k and then instead of omega I am doing an omega rho boundary of omega rho is a union of boundary of omega and s xi rho for example this is omega xi radius rho I am removing this so this is my domain where is the domain this is the domain so this domain has 2 boundaries one is this boundary and one is this boundary since Laplacian k is 0 for x different from xi now in omega rho there is no xi xi is taken out therefore this is 0 and hence this term drops out. So what we have is the first term on the LHS equal to this quantity and boundary consists of 2 parts so I have input at that one is boundary of omega other one is s of xi rho this is a sphere. Now let us look at this term and try to simplify this term because assertion 1 contains this term this term and not this term but a simplified version of this. So let us look at the second term let us compute the second term on RHS of this equation this equal to this is the first term here minus the second term so let us address each of them separately. Note that for x on the sphere s xi rho we have k x xi equal to psi of rho using this information on divergence theorem we get k dou n nu equal to k is psi rho so it comes out it does not depend on the integration variable because k is constant so the psi of rho that comes out and integral of dou n nu over s 0 this is where we apply divergence theorem and we get in terms of Laplacian so minus psi of rho integral over the ball Laplacian u dx the outward normal n on the sphere points towards its center xi. Let us see our picture this is our omega and inside that we have removed a ball our domain is really this one if you take a point here normal if you take this side it is the inside pointing normal so this is not the one so this is the one which is outside pointing outward pointing therefore this is towards the center of this ball also note the dou n k of xi is nothing but minus psi dash of rho holes at points on the sphere s of xi rho. Thus we get integral of u dou n k equal to minus rho power 1 minus d by omega d integral over the sphere u d sigma. So, on the last slide we have proved this equality thus the second term now is given by minus psi of rho integral over the ball of Laplacian plus rho power 1 minus d by omega d integral over the sphere of u d sigma. Since both u and Laplacian u are continuous at xi we have assumed u is c 2 of omega bar as rho goes to 0 we have psi of rho into integral over the ball of radius rho of Laplacian goes to 0 because modulus of psi of rho into this integral term is less than or equal to m times psi of rho into the volume of this ball. What is m? m is a bound for modulus of Laplacian u now psi of rho is like rho power d minus 2 whereas the volume of the ball is like rho power d therefore their product will behave like rho square. So, therefore as rho goes to 0 this term goes to 0 rho power 1 minus d integral of this sphere will go to the omega d into u of xi where omega d denotes the surface area of the unit sphere in rd. Please check these assertions by yourself. Thus we have the following convergence of the second term as rho goes to 0 this is the second term this goes to u of xi because the first term went to 0 second term went to u xi finally pass into the limit as rho goes to 0 in this equation we get this equation. This completes the proof of 1 u of xi equal to this integral minus this integral this is what is stated in 1. As mentioned before statement 2 follows immediately from statement 1. Statement 3 follows from statement 1 by taking u equal to phi which is c0 infinity of omega this completes the proof of the theorem. Remark this formula we have just proved this is assertion 1 gives a representation of the solution you want to know u of xi it gives in terms of this k Laplace in u is okay if you are solving Laplace in u equal to f this is known if Laplace in u equal to 0 this term is not there. So, this is a known term k is already known but this second term involves dou n u as well as u if you are solving Dirichlet problem u is known but this is not known if you are solving Naumann problem dou n u is known on the boundary but u is not known. Let us discuss this point a little bit of course this represents a solution if it exists of course we know that if solution exists it is going to be unique we already proved that so this formula is a representation for u of xi in terms of values of u and values of dou n u on the boundary of omega. However, for Dirichlet problem note that only the values of u are prescribed on boundary of omega that means only this term is known and this is not known and thus the formula given above is not useful for computing the solution note that the boundary values of u already determine a solution to Dirichlet problem and thus the quantity dou n u is not only not known it is already determined we now present two sample theorems without proof which justify the naming of k of xi as a fundamental solution. Theorem on logarithmic potential naming will be obvious once we state the theorem let fpc2 function defined on R2 having compact support define the logarithmic potential on R2 by u of xi equal to 1 by 2 pi integral over R2 ln of norm x minus xi fx dx. Then the following assertions can be proved of course we are not proving that is why I have stated as the following assertions can be proved logarithmic potential satisfies Laplacian u equal to f that means this formula is a solution to the Poisson's equation and u of xi goes to infinity as norm y goes to infinity. In fact, we have the following asymptotic behavior of the logarithmic potential at infinity u of xi is equal to m by 2 pi log norm xi plus big O of 1 by norm xi where m equal to integral of f over R2 which is a finite quantity because f is assumed to be compact support so integral is finite. Logarithmic potential is the only solution to Laplacian u equal to f having the asymptotic behavior as mentioned in two above. So, interpretation of potential a function u satisfying Laplacian u equal to f is said to be the potential due to the charge f in the context of electrostatics. Theorem on Newtonian potential let f belongs to c2 of R3 having compact support define the Newtonian potential in R3 by u xi equal to minus 1 by 4 pi integral over R3 of fx by norm x minus xi dx. Then the following assertions can be proved Newtonian potential satisfies Laplacian u equal to f u xi goes to 0 as norm xi goes to infinity. Newtonian potential is the only solution to Laplacian u equal to f that is in c2 of R3 and vanishes at infinity. So, let us summarize what we did in this lecture. Idea of a fundamental solution was introduced. Fundamental solution for Laplacian in Rd was obtained for d greater than or equal to 2. How solutions to Laplacian u equal to f may be obtained using fundamental solutions for Laplacian was mentioned the two theorems. In the next lecture we will discuss the role of fundamental solutions for Laplacian in determining solutions to Dirichlet boundary value problem. Thank you.