 Welcome to 11th lecture and from 11th lecture now we are getting into finite element analysis. There are two distinct approaches in finite element method. One is the variational approach which is basically energy minimization approach which is more sort of closer to the physics because you know we are calculating the energy and we are minimizing the energy and why we minimize energy basically because fields will generally, fields will always get distributed in a given problem domain such that energy content is minimized that is you know is one of the nature's laws. So in the fields will get distributed such that energy content is minimized. There is another approach weighted residual approach wherein we minimize errors because of approximation that we are doing. So we will see you know this weighted residual approach little later the first two set of lectures we will concentrate on this variational approach which is basically energy minimization. See here and remember we are taking again this parallel plate capacitor problem to understand what is this you know energy minimization and how energy gets how you know the field distribution corresponds to energy minimum condition. So here always remember that you know most of these problems that we are taking particularly in electrostatics you will find that we are not considering the charges on the plate because those sources because the source actually charge always remember in electromagnetic the basic entity that source is charge only that charge when it is moving then it is current that current can be DC or time varying but the basic source is charge all current. But here you know you may wonder you know in all these things we are not actually representing charge. But the charge is represented by the corresponding boundary condition here for the 100 volt and 0 volt. Now again for simplicity we are neglecting the end effect. So we are considering you know it is like in one of the previous lectures we saw in is it not we actually will terminate this whole geometry here like by this vertical boundaries and impose Neumann homogenous Neumann condition that is daba v by daba z equal to 0 in this case. Now actually you know here since it is a uniform field condition because end effects are neglected everywhere the field is uniform. So your electric field intensity is simply 100 volt by 10 meter because 10 meters is the gap between the two plates. So you have you know e a given by 10 volts per meter everywhere and we know the energy is you know integral volume integral half epsilon e square dv. Now see it is becoming easy if I had directly started with FEM then I would have to explain what is this energy. Now we have covered almost all background theory required for FEM and it will be easier for you to understand electromagnetic aspects of FEM. We have to concentrate more on the mass aspects of the FEM in the lectures that are going to follow. So energy is given by this and now here this dimension is shown here as x so you know you have in yz it is infinite in yz extent. Remember when we imposed here homogenous Neumann condition effectively you know you what you are doing is you are making the plate this plate infinite in extended z direction. In y anyway into the paper it is infinite. So effectively what we are doing it is this dv is dx into 1 into 1 like we saw in basics when we did 2D approximation we always said dx into 1 is it not per meter depth and also we saw the parameters are calculated per meter depth so here it is since it is you know you are assuming infinite extent in both x and y and z you have it is basically you are per meter depth in those. So that is why this will this problem will essentially reduce to 1D is it not is a 1D problem effectively and then we also know for this problem the solution is the equi-spaced lines the equi-potential lines will be equi-spaced is it not all these lines will be equi-spaced but you know let us assume so by our logic that we just you know saw the nature's law so this should lead to minimum energy condition right so we need to prove that this indeed equi-spaced equi-potential lines indeed lead to minimum energy condition so we will have a counter argument and then sort of prove it right. So let us assume that no these equi-space lines are is not energy minimum condition so this dotted line here shown suppose you know that line is representing 90% of voltage which is say 90 volts right and you know the so the solution is not equi-space line but one of the lines is you know not equi-spaced it is above the actual solution as but then now what you will see the energy in this only if you consider energy in this region up to this you know 80 volts you know energy up to 80 volts line in the previous case when it is exact solution the energy will be proportional to 10 square plus 10 square because e was 10 is it not 10 volts per meter so e square 10 square plus 10 square that will be proportional to 200 but in case of dotted line you will have you know assume that since this line is taken above in the top this region between the droplet and this dotted line the electric field intensity will be more is it not and that I have assumed that it is say 11 volt per meter and in this case between 80 volt and 90 volt now the electric field intensity will be lower which I am taking it as say 9 volts per meter now you can see you know 10 square plus 10 square is 200 and 11 square plus 9 square is 202 is it not so this that is why the dotted case would represent higher energy than the exact solution so in a way if we have proved that in this case the equiposive potential line equi-space equipotential lines lead to the minimum energy condition right so now actually if you take the other case where in which also we have been seeing previously lead to ground now here the equipotential lines are going to be like this is it not we have seen that now these lines are not equi-space the actual exact solution of this would corresponding will would correspond to equipotential lines would which you are not equi-space because the lines which are closer to the conductor they will be more closely spaced as compared to equipotential lines which are towards this ground electrode they will be sparsely spaced right so now here in this case the exact solution corresponds to this non-equi-space equipotential lines and in this case suppose you know you assume equi-space equipotential lines then in fact you will calculate higher energy than the exact solution which is non-uniform case and not equi-space line is this point clear right now you know here equi-space lines that will have to you know actually from the FEM solution you will have to work out then and that is not very straightforward to calculate it was very simple in case of parallel plate capacitor to prove but here it would be difficult but I hope intuitively you understood that you know why energy minimum condition is the exact solution from the previous you know capacitor problem now let us go further so we are considering again I am repeating it is a source free electrostatic problem because we are not considering charges on the plate right but that source of charges is being represented by corresponding bonding conditions on the top and bottom plate and the corresponding energy is given by what is written there now before going into you know the details of this how to solve this electrostatic problem let us see the two most standard partial differential equation pd stands for partial differential equation in electromagnetic so we will be basically considering three cases typically for low frequency electromagnetic when we are dealing with electrical machines and equipment one is the magnetostatic case del square a is equal to minus mu j where a and j are vectors now again you are very comfortable with this equation is it not Poisson's equations for magnetostatics where a is the magnetic vector potential j is the current density the second equation also we have seen you know we had some discussion also on this I have also explained you know this this is the induced eddy current term right because this dA by dt is the induced electric field intensity so sigma dA by dt is j induced right so that is why the units are matching because the unit is mu times j so these also mu times induced eddy current density is it clear okay so del square a will be equal to minus mu j if it is a static case if it is a general transient case this will be the case wherein this is the source source current and this is the this is representing the eddy current and I as I mentioned in that previous lecture when you bring this on this side this sign becomes plus as it it should be because source current and induced eddy current they should have opposite sign right but their angle will not be exactly 180 degrees mind that so that will depend because this will be all phasors and angle will get decided by you know many other considerations but maybe little later we will discuss this but in general the sign should be opposite and now if you actually are dealing with time harmonic case that means if your quantities are varying sign certainly with you know time then you can make convert d by dt as j omega and then you get this expression where now this a and j they are phasors are not you know represented by another symbol for sake of you know gravity and simply for simplicity right but remember this in this thing a and j they will be phasors now let us go further so now we have to minimize this now we are going back to electrostatic problem so first we will in this you know remaining course what we will do is we will start with electrostatic then we will go to magnetostatic field then we will go to time harmonic then we will go to you know axisymmetric we will get into permanent magnet we will go into transient then forces coupled circuit field and in that complexity we are going to proceed in this course so again you know always electrostatic is probably the simplest thing to analyze so we are starting with electrostatic so you know you are minimizing this and this we have already seen this for 1d electrostatic you get 1 into 1 so it is basically only d x now here for a standard you know capacitor problem we know we knew that the energy is this because it is related to you know capacitor and then half epsilon e square is the energy density right and that is why we started directly with you know assuming that this is the energy that has to be minimized but in general you know how do we sort of for a given partial differential equation we have to know its energy the corresponding energy to be minimized right so we will later on see how do we you know find out the corresponding energy functional that need to be minimized for each of these you know partial differential equations right so then we are coming to an important concept of what is known as functional now functional is basically is the energy so functional is nothing but the energy so in variational calculus we basically term energy as functional so functional f phi now this phi here is potential right that is the reason you know in the previous in the basics we you know call flux by psi because you know it will otherwise interfere with this in call of you know potential so here phi is the potential it could be in general electric potential magnetic potential of any any potential right it is generally it is a phi okay in this case if you are dealing with electrostatic it will be you know electric potential V okay so this is V for electrostatic now in this case this functional it can be defined in general of this form p1 to p2 that means this there is a one-dimensional domain so this is a one-dimensional domain so it could be is a one-dimensional domain so this is x and this is say phi right and this is a p1 this is p2 so this is a one-dimensional domain from p1 to p2 right and say potential may be varying between this like this it may vary in any way I just shown by a straight line but it is not it can vary depending upon the boundary conditions and the source condition right so here now for example so this functional energy can be in general represented by f of x phi phi dash now phi dash is the derivative right now here for example if you compare here this functional for electrostatic case here only what is that appearing here only V dash because e is nothing but dv by dx is it not so only V dash is or phi dash is appearing so this e is nothing but this phi dash so for this case is only you know it is a energies function of only phi dash but in general it will not be so it will be generally in the function of x phi phi dash for example in case of Poisson's equation you will get you know it has a function of phi also that we will see later so there will be some term in this expression which will be function of phi right so here so then you have one independent variable for this case and dependent variable is phi, phi is the potential right so you know you have this Laplace equation only phi dash appears in functional and as I mentioned Poisson's equation phi also appears so we will go further so this is what we said and then Laplace equation only phi dash appears in the energy functional but later on we will see Poisson's equation for PD for the corresponding functional for Poisson's equation we will have phi term also. Now we will go further and discuss more about this functional F and the corresponding variational calculus right now here we know total differential d capital F is daba F by daba x into dx daba F by plus daba F by daba phi into d phi plus daba plus daba F by daba phi dash d phi dash right but when it comes to variation in F in variation in F when we are doing we are not varying x so that is why we are saying del x is equal to 0 what it means we know that you know we want to minimize this energy which is represented by capital F that energy when we will get minimize when it will get minimize when we vary potential at each point and for one combination of potentials at various points we will get minimum energy. So what we are doing is in this figure now this one dimensional domain there are so many points x and there are so many points here at each of these points we will vary the potential values right. So at this point example for example we are varying potential from some initial guess value to something higher value so at this point it is varied like this. So at every point we are varying the potential but we are not varying x we are not varying potential as function of x so we are at every x we are varying potential right. So that is why here we are varying potential so that is given by del phi so there is variation we are doing variation in phi similarly as phi is varied phi will be a function of x so phi dash also will vary so again there will be variation in phi dash so as you vary potential there will be variation in potential derivative also but we are not varying del we are not varying x that is why here del x here will be 0 that is why this term is not there so f by x into del x this term will be 0 because del x is equal to 0. So this is the main concept in when we do variation. So again you know del phi is variation in phi at fixed x so this is the various phi distribution that we are attempting to minimize the functional f which is the energy. Now here let us come to this graph now this figure now here what you see is variation of f with respect to phi now remember this when we are saying this phi and this phi plus del phi these are set of values now suppose if there are say 100 points here say and we calculate the energy associated with you know 100 points so and that corresponding total energy what will happen is this phi will correspond to a vector of 100 values. So all those 100 values will decide the energy of that region one dimensional region so now if you change the potentials or at those 100 points you will get new set of set of values of potentials that will give another this kind of potential versus x and that may give some other energy value. So now when actually you are approaching the minimum energy point what will happen when you suppose for example this is the minimum energy potential distribution now if you change the potential values at all the nodes by some small value it is not going to lead to any appreciable change in the total energy content so that means we have reached the minimum that is the meaning of this statement del f capital F is equal to f phi plus del phi is my minus f phi is equal to 0 when this becomes 0 that means we are near the minimum energy point this will lead to 0 and that is the corresponding phi is the solution. Now you know let us expand this further so what does this mean f of phi plus del phi is nothing but f of x plus del x phi plus del phi phi dash plus del phi dash integrated over you know this domain with dx as you know integration distance minus this f phi as it is. Now again I just want to repeat f is just function which is function of x phi phi dash integral f dx is capital F so functional f capital F is function of function right so this is again to just highlight the difference between capital F and small f. Now this expression this equation if we further simplify this you know by using standard rule that f of x plus h is equal to fx plus hf dash x and there are remembering that there are two variables here this and this and this del x is equal to 0 so that is why it will be only f of x and then this because del x is equal to 0 so that is why here you know you get as written here. So now you have only two variables here this will get expanded to this plus this right by the formula that I just mentioned previously. So now this term can be written as this f by f of phi I am now just omitting this thing for simplicity and now this g remember this g is del phi is gx g as a function of x because this variation at every point is a function of x because here the variation is small here the variation is bigger here again it is becoming smaller. So the variation that we are making in phi is function of x right so that is why g is a function of x and that is nothing but del phi so this we are writing it like this and then this term we are you know splitting by using integration by parts and these are standard integration by parts and then this term first term there which is rewritten here that becomes 0 because at when that term evaluated at boundaries it will be this will reduce to 0 because g is 0 at the boundaries because potentials are already fixed at the two boundary points right. So this becomes 0 and then these two terms this term and this term these two terms are combined this is 0 so you get integral of this equal to 0 so when this is this integral is equated to 0 we are basically minimizing the energy right. So if this has to be 0 and if energy has to be minimum effectively this integrand has to be 0 because for any arbitrary gx g of x this whole thing can be 0 only if this is 0 right. So this is called as Euler-Langrange equation and this you bracketed term is called as Euler-Langrange expression right. Now let us see some you know application and example to understand what we just now saw in little bit more depth. Now let us again take a parallel plate capacitor problem and we know the functional for that is half integral epsilon E square dx right and is one dimensional that is why this dv reduces to simply dx into 1 into 1 right. So now E is nothing but dv by dx right and then that is why f becomes half the function becomes half epsilon dv by dx whole square right dv by dx square and then integral f dx is nothing but capital F. Now if we substitute now if we suppose we want to solve this capacitor problem and find out the distribution we will assume some potential distribution v right. If that potential distribution v is exact solution and we know it then when it is substituted in this Euler-Langrange equation the energy will be minimized because solution is known and it is exact. So when you substitute in Euler-Langrange equation you will get it equal to 0 and that is what so here we are assuming that this v is known exact solution is known and when it is substituted in this Euler-Langrange equation we will get on the right hand side 0 and then if that energy gets minimized in that way this v is known exactly. Now let us see when in Euler-Langrange equation when you substitute this f and then further simplify you will get this as d2v by dx square equal to 0. So this we are getting Laplace equation in 1d. So when this exact solution say was known to us and we substituted in Euler-Langrange equation we will have the minimum energy condition and therefore the right hand side of the equation will be 0 satisfying Euler-Langrange equation. This further leads to d2v by dx square equal to 0 which is Laplace equation. So again I want to highlight this here f was only function of v dash dv by dx but in general it can be function of xv and v dash. We will see such examples in some other cases later. Thank you.