 During the last lecture, we have been talking about the Poisson's and Laplace's equation, and we also discussed how to obtain capacitances of few geometries starting from Poisson's and Laplace's equation rather than from the intuitive nature of the potential which we had earlier. In this lecture, we will continue with the formal solutions of Laplace's equation which is the equation which is valid in regions where there are no charges. You recall that Poisson's equation which is del square of the potential is equal to the charge density divided by epsilon 0 with a minus sign. That was valid in regions where there are charge sources of charge present, but we will be talking about source free region namely charge density is equal to 0. Therefore, the equation that we are looking for is the Laplace's equation which is del square of phi equal to 0, phi is my notation for the potential. What we will do is to look at formal solutions of these equations in 1, 2 and 3 dimension. So, let us look at first the simplest of the problem. Namely, I look at the problem in one dimension for which this equation is d square phi by dx square is equal to 0, the simplest of the second order equation that you can think of. So, if the second derivative is 0, the first derivative d phi by dx that is equal to a constant. So, let us call it some m and phi itself is one more integration which gives me m x plus a constant equal to c. We all know that y equal to m x plus c is an equation to a straight line. So, as a result the equation that I have, the graph of the potential versus the distance is something like this. So, straight line, so notice one thing the solution of this equation as we will see it is a common feature with Laplace's equation is essentially featureless, the very monotonically increasing function in this case or decreasing function. And important point is that if you are looking at the value of the potential at any point x, since this is a straight line, what you could do is to find out for example, what is the value of the point x. You could find out what is the potential at let us say x plus a and at x minus a and you can easily show that the potential at the point x is the average of the value of the potential at a point x plus a and x minus a, a is absolutely arbitrary. So, of course it is very clear from the nature of the solution that there are no maxima or minima in the solution of this problem. We will see that this averaging is actually a common feature of the solutions. So, since it is an averaging I cannot obviously have the minimum or a maximum because then the value at a midpoint between maximum and medium, minimum cannot be actually an average. So, let us look at what happens in two dimensions. Once again the same statement is true namely the function which are solutions of Laplace's equations, they do not show any features, they do not show any maximum or minimum. Let me illustrate this with a very simple example. Let me look at a form of an equation which satisfies the Poisson's equation. For example, consider a function of x y which is simply a by 4 a is a constant into x square plus y square. Now, let us look at what is del square. Remember we are in two dimension. So, del square implies d square by d x square partial derivatives plus d square by d y square partial derivatives. So, therefore, my del square of phi in two dimension is simply equal to a. You can see it first derivative gives you 2 x a by 4 that is a x by 2. The second derivative will give me another it will reduce the power by 1. So, I will have a by 2 into x just a by 2 I am sorry just a by 2. And similarly, the second derivative of this will be another a by 2 and as a result the total second derivative is equal to a. Now, this is of course, not a Laplace's equation. This looks like a Poisson's equation. Now, this solution if you plot here is a genu plot of the function. You notice this function that it looks like a cup and so there is a minimum at the point x equal to 0, y equal to 0. And that is very obvious, because this is a positive function everywhere. And since the function is positive it is absolute minimum value can be 0. So, once you are at the once you are at the bottom then of course, the value of the function has to rise no matter which direction you go to. So, this is a function with a minimum. Now, consider for example, a slightly different function namely phi of x y the phi of x y is let us say instead of a by 4 x square plus y square I have x square minus y square. Now, you can immediately check that del square of phi works out to be equal to 0 which means this is Laplace's equation. Now, supposing you are to plot this function this is del square phi equal to 0 a by 4 x square minus y square. Now, what happens at the point x equal to 0 y equal to 0 is that a minimum the answer is now. Look at this structure here. Now, the point x equal to 0 y equal to 0 is what is known as a saddle point. You all know what is a saddle. A saddle is a seat on the back of a horse. Now, if you have looked at a saddle you will realize that in one direction the if you go in one direction that is along the back of the horse this function will go down, but on the other hand if you go in the other direction the function will go up. It is a saddle it is not a maximum because there are regions which where the value of the functions is greater than the value at x equal to 0. On the other hand it is not a minimum because there are regions for example, in this part where the value of the function is smaller than its value at x equal to 0 y equal to 0. So, once again that tells me that the Laplace's equation solutions at least from the examples that we have pointed out. They do not show any features it does not have a minimum it does not have a maximum end. So, we have seen that the averaging property works and things like that. A consequence of this type of thing that we are doing is what is known as an Ernst's theorem. Now, Ernst's theorem states that supposing you have a charged particle the theorem states that it is not possible for a charged particle to be an equilibrium static equilibrium only by action of electrostatic forces. There are other ways of keeping it in equilibrium, but if the only force that is there is an electrostatic force you cannot keep the charge in equilibrium. Now, why now look at this I know that if I have a charge q in a potential field v the energy of that will be equal to q times v. Now, if the energy is q times v and the charge is in a region which satisfies Laplace's equation because there are no sources in that region. So, I have del square v equal to 0 which implies that the del square of the energy function which is q times the potential that is also equal to 0. Now, if del square of v equal to 0 or del square of q v I am using the symbol v for both potential and phi also for potential should not cause any confusion at this moment. Now, this solution does not have a minimum at least we have seen just now. Now, if the energy function does not have a minimum then clearly the configuration cannot be stable it cannot be in stable equilibrium. So, this is known as Ernst's theorem follows from the fact there is no minimum in energy which in term follows from the fact that the potential function satisfies the Poisson's equation. Now, we will come back to solutions of Laplace equation in two dimension little while later, but at this moment we will jump to these solutions in three dimension in different types of geometries. The three types of geometry which we are concerned with primarily are the rectangular Cartesian for which the Laplacian equation is simply del square phi by dx square del square phi by dy square and del square phi by dz square this is Cartesian form. These spherical polar form depending upon distance r polar angle theta and azimuth phi is given by this expression here we will be talking later about this equation in detail. In cylindrical rho theta z coordinate as you know that the z coordinate is same as that of the Cartesian. So, therefore, the Cartesian expression remains the same and the other two are basically the polar in the x y plane and that is the way it looks like. So, let us look at the solution of Laplace equation in three dimension first we will take up the rectangular or the Cartesian coordinates and then we will look at the spherical polar. Now, in Cartesian coordinates we have seen that I have got del square phi by dx square plus del square phi by dy square plus del square phi by dz square equal to 0. We know that we would like to solve this problem subject to some boundary conditions. Now, let me let me say I have chosen a region of a rectangular parallel piped whose dimensions are length breadth and height are l 1 along the x direction, l 2 along the y direction and l 3 along the z direction on the six surfaces. I have certain values of the potential specified. So, these are my boundary conditions. Now, what we will do however is use the fact that the solutions obey superposition principle. So, instead of giving all the six conditions I will work out in detail the solution for the problem where on the five sides of the rectangular parallel piped the potential will be taken to be 0 if they are conductors it means they will be grounded and on the sixth side I will say that the potential function is known to be. So, let me state that in this way. So, I have this this is my x y and the z direction coming out of the plane of the picture and on five of these that is when x is equal to 0 and x equal to l 1, y is equal to 0, y is equal to l 2 and z is equal to 0 these five surfaces the potential value is 0. On the sixth surface which is at z is equal to l 3 the potential distribution as a function of x y is known to be. So, that is f x y how does one solve this problem. Now, supposing you have six different boundary conditions I could actually solve these subject to a given boundary condition on each one of them and then sort of make a superposition of these to get a general solution, but at this moment I am looking at a specific problem where on five surfaces the potential value is given to be 0 and on the sixth surface which is at z is equal to l 3 the potential function is known to be some given function f which is of course, a function of x and y because the z is constant there. Now, so let us look at this equation the method that one follows is what is known as separation of variables. The separation of variable works this way that this function phi is taken to be a product of three functions function capital X which depends upon x only function capital Y depends upon y only and function capital Z depends upon z only. Recall at this moment I do not know whether this trick will work or not, but if this trick works then I will fall back on the uniqueness of the solutions and say that look by this trick I have found the solutions and so therefore, that must be the only solution. So, let us look at this. So, you notice this that so the what will happen is look at this term d square phi by d x square. So, since capital Y and capital Z are constant I will get from this term y z times d square capital X by d x square plus here I will get x z d square capital Y by d y square plus x y d square capital Z by d z square that is equal to 0. Let me divide this equation by capital X y z that will give me 1 over capital X d square x by d x square plus 1 over capital Y d square y by d y square plus 1 over capital Z d square z by d z square that is equal to 0. Now, look at this equation in this equation I have got three terms the first term depending on X only, second term depending upon Y only, third term depending upon Z only and there are variations here with respect to X here with respect to Y and here with respect to Z and I am making a big demand that some of these three must be equal to 0. Now, since x y z they vary arbitrarily in this space this is too tall an order and the only way we can do that is that if each one of these terms is a constant subject to the fact that three of those constants coming from three equations become equal to 0. So, this is purely because of the fact that the variation of each is independent. So, what do I get? So, let me write then the first equation will be 1 over x d square x by d x square is a constant. So, let me take this to be equal to some let us take minus k x square times constant this k x square the this x is because I am getting this constant related you know associated with this x. So, this is a constant value let me take d square y over d y square is equal to minus k y square into y, but d square z by d z square then this is a constant then has to be related to these two constant in such a way that the corresponding k z square if I take it. So, I will take d square z by d z square is equal to plus k z square I will take z subject to the fact that k x square minus k y square plus k z square is equal to 0 where all that we know about k x k y and k z are that their constants. Now, let us look at what happens to the corresponding boundary conditions. Remember on five of the faces the potentials were 0. So, I can translate that and say that look since x the depends only on x. So, this implies that the capital X function at x equal to 0 must be 0 and it also has to be 0 at the value l 1 the same is true for y, y must be 0 at 0 and at l 2 about z of course, I have a slightly different story because the value of capital Z must be equal to 0 at z is equal to 0, but at z is equal to l 3 it must be equal to constant. The solutions of these equations for example, the first equation which I have taken as d square x by d x square is equal to minus k x square x and also we learn to all of us and the solutions of course, are capital X as a function of x is a sin k x x. What is that I have chosen these two surfaces the at x equal to 0 I have got the value of the capital X equal to 0 it is automatically satisfied because I have taken a sin function, but there is a one more demand we say that x at x equal to l 1 is also equal to 0. This implies that sin of k x l 1 is equal to 0 which immediately tells me that the k x values are equal to must be such that it must be multiple of pi over l 1 where m is an integer m is an integer it is plus or minus 1 plus or minus 2 etcetera. I cannot take m to be equal to 0 because that would mean that the function itself is 0. So, which is of course, not what we are looking for. Now, identical statement is true for the y function. So, y of y is equal to sin k y y the boundary condition at y equal to 0 is automatically satisfied and the boundary condition at l 2 tells me that this should be an integer n divided by l 2. So, what is so good the four on four sides the solutions have been satisfied. Now, once I have taken once I have taken k x square and k y square to be positive or k x square and k y square to be positive the k z square term which is associated with the z equation is a negative that its solution cannot be trigonometric function because that equation now is d square z by d z square is equal to plus k z square z and since this is positive value I know its solution will be exponential with real arguments. However, I also have one more condition to be satisfied and that is the value of capital Z. At z is equal to 0 is also 0. Now, I know they are not trigonometric function they are exponential function and. So, therefore, what I do is to choose this solution to be a hyperbolic sin function. So, we will take capital Z to be given by hyperbolic sin of k z z remember sin hyperbolic of z at z equal to 0 is 0. So, this is this is the solution I take is actually not equal to, but should be constant times that. Now, let us look at the k x k y and k z we have seen that k z square is equal to k x square plus k y square. So, that tells me that this k z is given by square root of k x square plus k y square and since k x is m pi over l 1 square and k y is n pi over l 2 square root. So, this quantity k z can be also written as k m n it tells me that in order to specify k z you need 2 indices m and n. So, therefore, my solution which is phi x y z is now a linear combination linear combination because m and n can vary from 1 to infinity each one of them. Coefficient c m n sin function of m pi over l 1 into x another sin function of n pi over l 2 into y and of course, sin hyperbolic of k m n z. So, this is my solution, but there is still a sixth boundary condition to be satisfied and this sixth boundary condition is that when z is equal to l 3 the value of the potential at every point on that plane that plane you remember z is given, but x and y varies and that is a known function f of x y. So, if you look at this then we get phi of x y and z is equal to l 3. This is simply substituting z is equal to l 3 into this equation, but this value is a function given function of x and y. So, this is equal to just to rewrite m n equal to 1 to infinity c m n sin of m pi over l 1 x sin n pi over l 2. Why sin hyperbolic k m n times l 3 that is your f x y. Now, I can determine this constant c m n by using the properties the orthogonal properties of the integrals. You notice that here I have given you an integral this is d x of sin m pi x over l sin m pi x over l where m prime is another index. Now, if you integrate this you will find only when m is equal to m prime it gives you l by 2 otherwise when you take the integral from 0 to l it gives me 0. So, that is why it is written as l by 2 times this is called a Kronecker delta m m prime that is equal to 0 if m is not equal to m prime and is equal to 1 if m is equal to m prime. Now, if you recall that integration that you notice that I come to this equation here I integrate I I multiply this both sides by sin m pi x over l 1 and n pi y over l 2 and integrate actually speaking what I do is to multiply with I should have said m prime n prime, but I know that unless m prime equal to m and n prime equal to n these c m n's will go away. So, this is what I get there this is what I get there and so that tells me that if I know the function f x y and I have been given everything else that I require there and in principle if I can do those integrations then in principle c m n these coefficients are known to me. So, let us let us take a very special case if I take a special case where I take this function f x y let us say is constant. Now, if you go back notice this that in c m n I have 2 integrals double integral over x and y, but if this f x y is constant then of course, these integrals become 2 product of 2 different independent integrals and sin integrals are very trivial integrals to be done. So, I use the fact that integral of d x sin m pi x over l is l by m pi cosine m pi x over l and the limits of integrations are from 0 to l 1. So, for d x integration this is l 1 and that is l 1, but this is a generic relation. Now, in the upper limit I have cosine of m pi I know that is nothing but minus 1 to the power m lower limit is of course, 1. So, this integral is l by m pi minus 1 to the power m minus 1. This tells me that this has a value of which is 0 if m is even because then it becomes 1 minus 1. If m is odd I get a minus 2 there because minus 1 to the power odd number is minus 1 and I also have another minus 1 there. So, I get so the result of this equation is it is equal to minus 2 l by m pi if m is odd and is equal to 0 if m is even. So, if you look at this expression for c m n is l by l 1 l 2 etcetera and I have this f x y which is now constant and product of 2 integrations and each one of the integral I know how to do it. So, if you plug it in you find this to be equal to 16 over m n pi square 1 over sin hyperbolic k m and l 3 and this will be non 0 only if only if neither m nor n is even because if any one of them is even because it is being multiplied they into corresponding integral is 0. So, when both m and n are odd c m n is non 0 in this case. So, that essentially completes our solution in the rectangular coordinates and with this we will go over to the solution of Laplace's equation in spherical coordinates. Spherical coordinates Laplace and form is a little complicated, but it turns out that the principle of separation of variable that we talked about still works here though with a slight bit of a modification. So, let us look at what does it get me first my del square of phi here I will be a little careful in writing the potential function phi because the azimuthal angle is also denoted by phi. So, this is given by 1 over r square d by d r of r square d phi by d r is the radial part polar the angle part is r square sin theta d by d theta of sin theta d phi by d theta. And finally, I have got 1 over r square sin square theta d square over d phi square this is my azimuthal angle of phi that is equal to 0. Now, what we will do is we will have the same trick that is look at the possibility of separation of variable in this problem. So, we write capital phi as product of 3 functions capital R which is a function of r capital P which is a function of theta and let us write capital F which is a function of phi. Now, if you plug it in let us see what you get. So, first is of course, I have got 1 over r square d by d r of r square d r by r and P times F will come out because they do not depend upon r. This one is similarly 1 over r square sin theta d by d theta of sin theta and this time it is d P by d theta and r and F will come out and 1 over r square sin square theta d square over d phi square of F and this time r and P will come out this is equal to 0. Now, we do the same thing again that is divide this equation by r P F and the other thing that you notice is there is a 1 over r square all over and. So, therefore, I can take that r square to the right hand side it becomes 0. So, I am left with the equation which looks like the following. I get 1 over r d by d r of r square d r by r square d by d r plus 1 over sin theta times P d by d theta of sin theta d P by d theta plus 1 over F sin square theta d square F by d phi square that is equal to 0. Now, here the first thing that we need to do is to realize that in this equation in this equation I can take this sin square theta multiply the entire equation by sin square theta and realize them these two equation these two terms the radial term and the polar term they are both theta appears there on the other hand this will give me just 1 over F d square F by d F square. So, let me let me do that. So, let me multiply this with sin square theta by r d by d r r square d r by d r plus sin theta sin theta by sin theta by sin theta by P d by d theta sin theta d P by d theta this is equal to minus 1 over F d square F over d phi square. Now, I give the same argument again the left hand side is a function of r and theta and small r and small theta varies of r. On the left hand side the right hand side however depends only on phi and we are expecting this should be equal to that that is not possible unless each one of these terms happens to be a constant. So, let us let us write that constant to be equal to some quantity like m square. Now, I can I am now in a position to solve what is known as the azimuthal equation. The azimuthal equation then becomes d square F over d phi square is equal to minus m square F. We know that the solution of this equation is e to the power plus or minus i m phi. What is m? The value of m can be fixed by saying that if phi increases by 2 phi then this I am back to the same point if the azimuthal angle changes by an amount 2 phi I am back to the same point. So, my value of F must be cyclic in the sense whatever is the value of F at some value of the azimuthal angle phi that must be having the same value when the azimuth increases by an amount 2 phi which is possible if e to the power plus or minus i m into 2 phi is equal to 1. So, if e to the power i m into 2 phi that is equal to 1 then it is possible which tells me that m is an integer positive or negative integer. So, in this equation my m is an integer that has been worked out. Let us look at the next equation now having done that identified m. So, I will now write the equation in this way remember I had divided by sin square theta, but now I have brought the sin square theta to wherever it is and I have written 1 over r d by d r of r square d r by d r is equal to 2 terms one is this term which was already there and the other one is m square by sin square theta. This has made it possible for me to separate r and theta the polar separation. So, therefore, once again with the type of argument that we have been given each one of these terms must be constant. I do not know what that constant is, but anticipating some result let me put that constant to be equal to l into l plus 1 of course, I have not yet said what is l. So, therefore, I can write my constant anyway I like. So, I have got now 2 equations. So, the main equation is the polar equation is 1 over sin theta d by d theta sin theta d p by d theta plus this l into l plus 1 minus m square by sin square theta into p equal to 0. You can simplify this equation by making a substitution of variable mu is equal to cos theta. If you do that mu is equal to cos theta d mu is equal to minus sin theta d theta. Now, since I know that the polar angle goes from 0 to pi the mu which is taken to be equal to cos theta goes from minus 1 to plus 1. Straight forward to check that this equation can now be written in terms of the new variable mu like this. We will not actually be trying to solve this equation, but it turns out that this equation is well known in the field of differential equation. The field of differential equation and the solutions are known to be associated Legendre polynomials. So, this is an equation whose solutions people know. We will be talking about the nature of solution little later, but let us look at that. So, this is called associated Legendre polynomials. So, my part of the difference of the Laplace's equation now has an e to the power i m phi times the associated Legendre whose which are characterized by l in addition to m. Now, incidentally there are two types of solution there. One is written as p l m another is written as q l m. Now, q l m goes to infinity to infinity as mu goes to plus or minus 1 p l m does not, but p l m also goes to infinity that is it diverges as mu goes to plus or minus 1 unless l happens to be an integer. So, therefore, I take the integer in that the integer l the l to be an integer and then the solution are given by p l m which are of course, functions of cos theta and of course, you have to multiply it with e to the power plus or minus i m phi. Let us take a simple specific case supposing my solutions do not have azimuthal dependence. What it means is my problem does not have a dependence on the azimuthal angle in other words there is azimuthal symmetry in the problem. Now, if there is azimuthal symmetry in the problem the solutions cannot depend on phi and the only way to do it is to take the m to be equal to 0. Now, if you put m to be equal to 0 in the equation of associated Legendre polynomials the equation that I get is d by d mu 1 minus mu square this time I have removed the l m index d p by d mu plus l into l plus 1 into p equal to 0. This is known as the equation for Legendre polynomial the previous one is called associated Legendre polynomial. So, the Legendre polynomials are essentially special cases of associated Legendre polynomial having m is equal to 0. Now, the solutions of these Legendre polynomials are also well known and the you can actually check very easily the solutions. It happens to be a polynomial in mu which is a polynomial in cosine of theta. For instance you will find that the zeroth order polynomial is 1 it is very straight forward to check that if p is equal to 1 the left hand side is identically equal to 0 because d p by d mu. If my if l is equal to 1 l is equal to 0 p 0 cos theta is 1. If l is equal to 1 then of course, this becomes 1 into 1 plus 1 which is 2 and you can check that p 1 of mu is nothing but mu itself you can that is very fairly straight forward because if p is mu d p by d mu is 1 I have got a 1 minus mu square there. So, when you differentiate this with respect to mu I get a minus 2 mu there. So, therefore I get a minus 2 mu there and here I got l is equal to 1. So, it is 1 into 1 plus 1 which is 2 and p is mu. So, therefore this equation is satisfied and likewise you can find that p 2 cos theta is half into 3 cos square theta minus 1. So, this is the spherical polar equation. So, what have we seen? We have seen that when I have azimuthal symmetry my solutions are given by the angular part of the solutions are given by Legendre polynomial. I have not yet touched the radial equation, but in the remaining time let us look at what the radial equation looks like. If you recall that the radial part was 1 over r d by d r r square d r by d r that was equal to a big expression with angular part and we put both of them as l into l plus 1. So, this is the equation split of this equation you get a equation like this. This is actually a very equation which is very easy to solve you can check if you take capital R to be going as R to the power n then the equation is satisfied and the value that n takes is l or l plus 1. So, as a result the complete solution of this equation which whose properties we will be talking about in the next lecture is given by a which is a constant R to the power l plus b by R to the power l plus 1 into p l cos theta. Next lecture we will look at some applications of this solutions in three dimensions.