 Welcome to tutorial 1 on Laplace equation. So problem 1 is let D of 0, 2 denote the disc of radius 2 centered at the origin and S of 0, 2 is the circle of radius 2 with center at the origin. Consider the Neumann boundary value problem Laplace in U equal to 0 in the disc and the normal derivative is prescribed as alpha x square plus beta x plus gamma for x y belonging to S of 0, 2 where alpha, beta, gamma are real numbers. Now if the BVP admits a solution U which is C2 of D0 to closed disc that means C2 of closure of this open disc D0 to then find a relation among alpha, beta, gamma that must be satisfied. So this is an application of a lemma that we have seen in lecture 6.1. Recall from lecture 6.1 for the Neumann boundary value problem Laplace in U equal to f in omega dou N U equal to g on boundary of omega to admit a solution the data f and g must be compatible. What is that? Lemma let f belongs to C of omega bar and g belongs to C of boundary of omega. If U belongs to C2 of omega bar solves the Neumann boundary value problem which is here then necessarily this integral of f over omega is equal to integral of g over boundary of omega. So in the present problem f equal to 0 and g equal to alpha x square plus beta x plus gamma. So if alpha, beta, gamma are such that this integral over boundary of omega of alpha x square plus beta x plus gamma equal to 0 does not hold then the given problem does not admit a solution in C2 of omega bar. But we are given that the boundary value problem has a solution U which is in C2 of omega bar therefore this condition must be satisfied. So let us compute this integral if this does not hold then the given problem does not admit a solution U belonging to C2 of omega bar. Remember here that a solution of the Neumann problem is required to be only C1 of omega bar. But we have proved this compatibility condition and the assumption that U is C2 of omega bar that is the reason why we are assuming that U is in C2 of omega bar. So let us compute the integral s of 0 to alpha x square plus beta x plus gamma d sigma of x gamma. So x equal to 2 cos theta y equal to sin theta but y does not appear in our integral. So that will make this integral to be 0 to 2 pi alpha x is 2 cos theta so x square is 4 cos square theta plus beta times 2 cos theta plus gamma into the integration is 2 d theta 2 is the radius of the circle. If the radius is r it will be r d theta. So this is nothing but there is 4 alpha under the 2 here. So 8 alpha 0 to 2 pi cos square theta d theta plus 4 beta 0 to 2 pi cos theta d theta plus 2 gamma 0 to 2 pi d theta. So that is equal to 8 alpha times pi plus 4 beta times 0 because average of cos theta on its period 0 to 2 pi is 0 plus 2 gamma into 2 pi. That is nothing but 8 alpha pi plus 4 gamma pi which is equal to 4 pi if I take common I get 2 alpha plus gamma. So therefore if you belong to C2 of omega bar is a solution then definitely we must have 2 alpha plus gamma equal to 0 is a necessary condition. So let us move on to problem 2 invariance of Laplacian under translation and rotation of coordinates. Inverience of Laplacian under rigid body transformations. What are rigid body transformations? These are the maps x going to qx plus c where q is a rotation matrix and c is a fixed vector in rd. In other words first x goes to qx that means it is rotated and then you are translating the resultant with a c. So that is the rigid body transformations. So translations fix a c in rd x going to x plus c is a translation. Let us see that the Laplacian is invariant under this coordinate chain transformation. So here x is nothing but x1, x2, xd there is an element in rd. So let us denote the new coordinates as y. So actually we should be writing y is equal to some function of x which is x plus c. Since we are enough experienced with change of coordinates already I drop writing this phi and I write y equal to x plus c. So let us draw one picture x1, x2 and you have a function u of x1, x2 and here you have this is y1, y2 it is a point c in your this coordinate system. So it has moved to c, origin has moved to c and the mapping which connects this is x going to x plus c and the inverse mapping is y going to y minus c. So let us write u of x bar equal to w of y bar. So in terms of this new coordinates let us write the function as w and this is a relation between them. So actually this is w of x plus c and similarly we can write w of y is equal to u of y minus. So let us compute the derivatives and check that the Laplacian is invariant under change of coordinates which is occurring by translations. So let us compute dou w by dou y1 at a point y that is dou u by dou x1 at the point y minus c. Similarly dou w by dou yd at the point y is dou u by dou xd at the point y minus c. Now second order derivative is dou 2w by dou y1 square at the point y is once again dou 2u by dou x1 square at the point y minus c and same is true for all second order derivatives. Now let us write Laplacian these how we denote this Laplacian in y coordinates that means let us write this acting on w that means it is dou 2w by dou y1 square at y summed up to dou 2w by dou yd square. This by the above computations is nothing but dou 2u by dou x1 square I will write the argument at the end dou xd square at the point y minus c which is nothing but what is y minus c is x so that is nothing but Laplacian in the x coordinate acting on u. So what we have got is Laplacian in the y coordinates this operator is same as Laplacian in x coordinates that is why Laplacian is said to be invariant. Now let us show that Laplacian is invariant under rotations in the plane. So rotation in the plane angle theta say counter clockwise does not matter but the matrix that we are going to write represents such a rotation which is in the anticlockwise direction. So let us write the change of transformation cos theta minus sin theta sin theta cos theta acting on x1 x2 how do you check it is a counter clockwise rotation put 1 0 and see what you get put 0 1 and see what you get for example if this is my x1 x2 direction the new coordinates are like that so this is an angle theta this is y1 and so if I have a function going out of this as u as before under w like that and from here to here we have this transformation y equal to let us write that matrix q times x. Let q denote this matrix which is here cos theta minus sin theta sin theta cos theta. So then we may write u of x1 comma x2 equal to w of y1 comma y2. We are going to differentiate this compute the Laplacian in both the coordinate systems and show that Laplacian is invariant. So u of x1 x2 y1 y2 where y1 y2 are given in terms of x1 x2 by the matrix. So y is equal to cos theta minus sin theta sin theta cos theta acting on x1 x2. So this is y1 y2 okay now let us differentiate. So dou u by dou x1 at the point x1 comma x2 by chain rule dou w by dou y1 at the point y1 comma y2. So y1 y2 is a shortcut for writing this entire expression into derivative of y1 with respect to x1 which is cos theta plus dou w by dou y2 at the point y1 comma y2 into derivative of y2 with respect to x1 is sin theta. If we expand this what you get is cos theta into x1 minus sin theta into x2 and second component is sin theta x1 plus cos theta x2. So similarly dou u by dou x2 I do not write arguments but let me now we get minus sin theta plus dou w by dou y2 into dou y2 by dou x2 which is cos theta. So now I write the expression for the second order derivatives please check it dou 2 u by dou x1 square at a point x1 comma x2 is equal to dou 2 w by dou y1 square at the point y1 comma y2 into cos square theta plus 2 dou 2 w by dou y1 dou y2 at the point y1 comma y2 into cos theta into sin theta plus dou 2 w by dou y2 square into sin square theta. Similarly dou 2 u by dou x2 square turns out to be dou 2 w by dou 1 square into sin square theta minus 2 times dou 2 w by dou y1 dou y2 into cos theta sin theta plus dou 2 w by dou y2 square into cos square theta. So, summing these 2 equations what we get is Laplacian in x coordinates on the left hand side equal to Laplacian in y coordinates of w on the right hand side. So, that is Laplacian in x is Laplacian in y. That means Laplacian is invariant under rotations in the plane. So, let us look at rotations in Rd, d not necessarily 2. So, let y equal to qx, q is a rotation matrix. So, q satisfies this relation q transpose q equal to q q transpose equal to the identity matrix. So, as before u of x equal to w of y we compute Laplacian of u and Laplacian of w with respect to x and y coordinate system and check that they are the same. Let us express yi the ith coordinate of y that is qx and then ith coordinate of that. So, which turns out to be summation qij xj, j equal to 1 to d. It is very important that we get used to summation notations like this because once dimension is above 2 or 3 it is almost impossible to write in a simpler manner. Therefore, we must get used to the summation. If you have any doubt you just try for y1, y2, y3 and you can arrive yourself at this formula. As I told before we should have ideally written yi is equal to eta i of x which is equal to this expression here. But since we are enough experience we are not writing that. So, let us compute the derivatives dou u by dou xi at a point x. So, that is dou w by dou yj dou yj by dou xi that is dou by dou xi of yj what is yj from here this expression is for yi. So, yj means we have to change this notation because this is anyway summation. So, let us write yj let us write k equal to 1 to d qj k xk that is why this is called dummy index because you can just change j to k or whatever you want because it is summed up. So, dou by dou xi of yj what is yj summation k equal to 1 to d qj k xk and then we have to sum over j that is the chain rule. So, please do this computation by yourself and then only proceed. So, this quantity is nothing but qij are constants. So, dou by dou xi of this entire quantity essentially is the same summation k equal to 1 to d qj k dou xk by dou xi dou xk by dou xi is 1 if k equal to i otherwise it is 0. Therefore, this quantity is nothing but it is nonzero only if k equal to i in which case you get qji. Therefore, this is equal to summation j equal to 1 to d qji dou w by dou yj at this point which if you want you may see it as this q transpose grad w with respect to y and ith component that is precisely dou u by dou xi and this is of course true for all i. So, let us compute the second derivative dou 2 u by dou xi square at the point x that is equal to summation j equal to 1 to d qji are constants and we have to write dou by dou xi of this quantity dou w by dou yj at y but y is qx by chain rule it turns out that this is equal to j equal to 1 to d summation n u summation will come q equal to k equal to 1 to d of dou 2 w by dou yk dou yj at the point qx are equivalent to y into qki qji these expression you will get but when we are interested in the Laplacian we must put a summation here from i equal to 1 to d and here i equal to 1 to d. There are a lot of summations here but they are all finite summations therefore we can easily interchange the order in which we sum what we do is we take the summation i equal to 1 to d and sum these up by freezing k and j. I will write this again on the next slide so thus what we have is summation i equal to 1 to d of dou 2 u by dou xi square at x the Laplacian and that equal to j equal to 1 to d k equal to 1 to d then we have dou 2 w by dou yj dou yk y and summation i equal to 1 to d qki qji we know something about this what is that we know that q q transpose equal to q transpose q equal to identity. So, this quantity i equal to 1 to d of qki qji is nothing but q q transpose this matrix and the kjth entry in that in that matrix that means an element which is on the kth row and jth column but q q transpose is identity. So, identity matrix and kjth entry this is a notation that we normally use delta kj it stands for 1 if k equal to j 0 if k is not equal to j therefore what we obtain at the end is summation i equal to 1 to d dou 2 u by dou xi square at the point x equal to so therefore this summation survives only if k equal to j therefore this summation will just become one summation with k equal to j you could use either k or j does not matter. So, k equal to j in particular means it is dou yk dou yk that means 2w by dou yk square at the point y which is nothing but Laplacian in the x coordinates of u at the point x equal to Laplacian in y coordinates w at the point y. In other words Laplacian x Laplacian y are the same so Laplacian is invariant under rotations in Rd. Let us move on to problem 3 solve the Dirichlet boundary value problem on annular region x square plus y square is between 1 and e square in other words look at circle of radius 1 this is the inner boundary of the angular region and this one with the radius so this is our region omega here x square plus y square will be bigger than 1 and less than e square and u is given to be 0 on the boundary so u is equal to 0 here u equal to 0 here also so the idea is to use polar coordinates and look for radial solutions. So, let us compute Laplacian in polar coordinates first so x equal to r cos theta y equal to r sin theta so w of r theta equal to u of xy but actually x equal to r cos theta and y equal to r sin theta so it is u of r cos theta r sin theta. So, let us compute the derivatives dou w by dou r at the point r, theta that is nothing but u x at r cos theta r sin theta derivative of x with respect to r which is cos theta and u y at the point r cos theta r sin theta and derivative of y with respect to r which is sin theta so this is dou w by dou r. So, if we compute dou 2 w by dou r square we get u xx I am not writing the argument it is r cos theta r sin theta into cos square theta plus 2 u xy sin theta cos theta plus u yy into sin square theta. This is dou 2 w by dou r square let us compute dou w by dou theta at the point r, theta that will be u x into minus r sin theta plus u y into r cos theta and dou 2 w by dou theta square equal to u xx into r square sin square theta plus 2 u xy into minus r square sin theta cos theta plus u yy into r square cos square theta. Let us sum this dou 2 w by dou r square plus 1 by r dou w by dou r plus 1 by r square dou 2 w by dou theta square. That will turn out to be dou 2 u by dou x square plus dou 2 u by dou y square. So, this is the Laplacian in xy coordinates and therefore this is the Laplacian in r theta coordinates. So, what is that Laplacian r theta coordinates acting on a function w of r theta is dou 2 w by dou r square plus 1 by r dou w by dou r plus 1 by r square dou 2 w by dou theta square. So, let us write the boundary value problem in the polar coordinates. Laplacian u equal to 1 has now become let me write w r r plus 1 by r w r plus 1 by r square w theta theta equal to 1 w of 1 comma theta equal to w of e comma theta is 0. So, let us look for a solution that does not depend on theta. If we are successful we will end up with a solution to the boundary value problem and we know that boundary value problem has a unique solution. Therefore, the solution we have found is a solution that is the idea behind looking for a solution which does not depend on theta. In case we are successful it is fine if not we have to admit theta dependence and try to solve again. So, w r r plus 1 by r w r equal to 1. So, this equation can be rewritten as r w r r plus w r equal to r that is if and only if r w r dash dash is derivative with respect to r equal to r. So, that tells us we can solve for r w r which is r square by 2 plus a constant k and that implies that w r is r by 2 plus k by r and from here we can again solve for w as a function of r only it does not depend on theta. So, I do not mention the theta dependence here. So, I am integrating the right hand side what we get is r square by 4 plus k log r plus another constant k dash we have to determine what are k and k dash using the boundary conditions. What are the boundary conditions we have w of 1 equal to 0 and w of e equal to 0. So, w of 1 equal to 0 will give us 1 by 4 plus k dash equal to 0 this will give us this and this will give us e square by 4 plus k plus k dash equal to 0. So, solving this system of equations we get k dash equal to minus 1 by 4 k equal to 1 minus e square by 4. Therefore, w of r equal to r square by 4 plus 1 minus e square by 4 into log r minus 1 by 4. So, in terms of x and y coordinates u of x y is equal to x square plus y square minus 1 by 4 plus 1 minus e square by 8 into log x square plus y square. So, it is not necessary to resort to separation of variables method all the time one could also use separation of variables to solve this problem. So, maybe that you can try let us move on to problem 4 let you be the solution to the Dirichlet boundary value problem in the unit disk a plus mu equal to 0 in the unit disk and on the boundary of the unit disk u is given to be 3 plus 2 x 1 plus 2 x 2 and find u of minus half comma minus half. So, how will you find one option is to explicitly solve this boundary value problem and then evaluate at minus half comma minus half another option is to use Poisson's formula indeed we are going to do that for xi in d of 0 comma r u of xi where u is a solution to this boundary value problem Laplace mu equal to 0 in the disk and u equal to g on the boundary that is given by Poisson's formula u of xi equal to r square minus norm xi square by 2 pi r into integral over the circle g x by norm x minus xi square d sigma in the given problem r equal to 1 g equal to 3 plus 2 x 1 plus 2 x 2 and xi equal to minus half comma minus half. So, u of minus half comma minus half equal to because what is the norm of minus half comma minus half we need norm square actually so that is equal to half. So, what we get is 1 minus 1 by 2 divided by 2 pi into integral s of 0 1 3 plus 2 y 1 plus 2 y 2 divided by norm of minus half comma minus half minus y 1 comma y 2 square d omega half. So, let us evaluate what is the denominator in this integrand so that is norm minus half minus half minus y 1 comma y 2 square is y 1 plus 1 by 2 whole square plus y 2 plus 1 by 2 whole square which on simplification you will get 3 plus 2 y 1 plus 2 y 2 by 2. Therefore, u of minus half comma minus half is nothing but 1 by 4 pi integral of s of 0 1 into 2 d omega of y. I have chosen simplest of the functions as the Dirichlet data on the boundary that is why it became very simple is 1 by 4 pi into 2 into the perimeter of unit circle which is 2 pi so that is equal to 1. So, let us move on to problem number 5 it is a Dirichlet problem on a rectangle. So, we have to solve Laplace equation in this rectangle and we are given boundary conditions only one of them is taken to be nonzero in case many of them are nonzero we have to split into sub problems where only one data is nonzero rest of them are 0 solve them and using linearity we can superpose and get a solution to the entire problem. So, let us solve this simpler problem we are going to use separation of variables method to solve this BBP we are experts therefore I just quickly go through the computations. So, we substitute u equal to x of x into y of y in the Laplace equation and then rearrange in the terms we get this now we observe that the LHS depends only on x RHS is a function of only y and that is possible if and only both of them are separately equal to a constant function. So, we put lambda here and from here we get the two ODE's we have to get the boundary conditions for these ODE's using the given boundary conditions we get this because we are not interested in finding solutions where either x is identically equal to 0 or y is identically equal to 0 because it is of no help in trying to obtain solution to the given boundary value problem. So, because of that we end up with these conditions for x and y. Now, we see that for x we have two boundary conditions second order equation so we will start solving for x and find out those lambdas which admit non-trivial solutions and for those lambdas we solve for y use this initial condition and then multiply them and propose a superposition as a formal series solution. So, this boundary value problem for x has non-trivial solutions if and only if lambda equal to minus n square for some natural number. Therefore, lambda n equal to minus n square are eigenvalues and corresponding eigenfunctions are sin nx. Now, for each n you solve this problem for y where lambda is equal to lambda n equal to minus n square solution is given by this up to a constant multiple. Now, for each n this product x and x into y and y solves the Laplace equation. So, therefore we propose a formal solution to the boundary value problem as a superposition of un's like this. Now, what is remaining is to find this bn and we have one more boundary condition which we have not used using that we will try to find bn and that is ux0 equal to fx. So, we get fx equal to this putting y equal to 0 this reduces to this. So, we need to find bn so that fx is given by such a series. So, therefore choose cn to be the Fourier sin coefficients of f you get bn equal to cn by sin hyperbolic Na. If f is given by this series then cn's are given by this formula we have already done this Fourier sin series earlier in the context of wave equation. So, please do the computations. Therefore, the formal solution is given by this expression. So, separation of variables method in principle you should be able to solve any problem because a basic idea is the same. Thank you.