We have been discussing approximate methods for analyzing structural dynamic problems as a build up to the development of idea of finite element method, so in today's lecture we will start talking about finite element method what are the basic motivations and start analyzing some simple problems, so let's first quickly recall what we have been doing, so with reference to a problem of beam vibration this is a governing equation for Euler-Bernoulli beam with specified initial conditions and appropriate geometric and natural boundary conditions, so the general format of finding approximate solutions is to seek the unknown dependent variable in terms of a series like this where phi N of X are known as trial functions which are to be chosen by the analyst and A N of T are the unknown generalized coordinates to be determined through an analysis, so the basic idea here is to we will substitute this assumed solution into the governing equation and we obtain an error or a residue term and the whole idea of developing approximate method is to drive some measure of that residue to 0 or as small as possible and use that criteria to select this generalized coordinates, so this class of methods are known as method of weighted residuals, so they lead to a discretized set of equations here A is a N cross 1 vector, M CK are all N cross N matrices, P of T is a generalized force which is N cross 1, we also discussed how to obtain initial conditions A of 0 and A dot of 0 in terms of specified initial displacement profile and velocity profile, now basically what does method of weighted residuals achieve? A partial differential equation governing the system behavior is now replaced by a set of ordinary differential equations with a view to obtain an approximate solution and that approximate solution is encapsulated in this representation. So we discussed several criteria for minimization of the residue, some measure of residue, so we discussed method of least squares, collocation method, galerkin method, subdomain collocation method, petrogalerkin method, etc. In all these methods the format of implementing the method of weighted residuals share certain common features, if we look at the way we obtain equations for A N in least squares approximation we minimize the total mean square error and we get a residue term multiplied by another function, so here also in collocation also we get the residue term multiplied by a weight function, so this is common to all these methods, so the format is therefore we have the residue term and we select N weight functions and write these equation for N equal to 1 to capital N thereby leading to a set of ordinary differential equations for the unknown generalized coordinates. Now the implementation of this method therefore requires choice to be made on two quantities namely trial functions and the weight functions, so if you look at the residue which will be equal to this term minus F of X, T it has fourth derivatives terms present in this example, so consequently the trial functions in this example need to be such that they are differentiable about to fourth order, but whereas the weight function does not, we need not impose any such requirements on the weight function. Now this format is known as weighted residual format or the weighted residual statement for the given problem, the original partial differential equation is called the strong format of the equation, in the so called weak formulation what we do is we want to now weaken the sense of continuity demand on the trial function and pass on some of that to the weight function, so that is achieved by integrating this equation in parts, by parts thereby what happens is the continuity requirements on trial functions come down and whereas on the weight function it goes up and the requirements will be shared equally between the weight functions and trial functions which the user anyway has to choose, so that approach is known as the so called weak formulation of the governing equations. We also discussed another method known as assumed mode method which is suitable for structural dynamic applications, you can assume the solution as in this form and directly work with the governing Lagrange equation to obtain the governing equations for A of T, so we have discussed all these methods and what are the limitations of these methods, now in all these methods we are representing solution in this form and this phi N of X are global in nature that is they are valid for all X belonging to 0 to L, so for simple line geometries as we have been discussing in the example so far the choice of this trial functions is reasonably straight forward we can make those choices with reasonable effort, but however if you are dealing with more complicated geometries or structures built up of line elements and so on and so forth developing global shape functions is not an easy exercise. Another problem with this representation is these generalized coordinates A N of T, N equal to 1 to N do not have a direct physical meaning, and we would like to deal with generalized coordinates if we could where these coordinates have some direct physical meaning. Now so to elaborate on what I said if you consider a three dimensional truss structure like this or a frame like this it's very clear that developing an exact solution for dynamics of this type of structure using by writing the governing partial differential equations for every member and establishing what happens at joints etc etc you are unlikely to succeed even for this simple example it will not be a straight forward exercise, so exact solutions are difficult if not impossible. Now approximate solutions using global trial functions are again difficult because it is very difficult to set up trial functions for this type of system similarly for this type even for this simpler system it is very difficult to set up the globally valid trial functions. As I again said the generalized coordinates even in such these representation do not have direct physical mean, now the finite element method basically aims to overcome these limitations. So again let's now start by making some observations and some general remarks what are the motivations for developing finite element method and what are the generic features before we get into specific details, so let me again summarize so the class of problems which can be solved exactly is limited, of course the exact solution have their own value when they are possible they serve as benchmarks against which approximate solutions can be validated therefore study of exact solutions has a place in our studies, on the other hand approximate solutions which imply global shape functions are possible for a limited class of problems. Again we would like to work with methods in which the generalized coordinates have direct physical meaning, unknown coefficients in a series expansion in terms of global trial functions do not serve this purpose, for most practical problems of engineering interest neither exact solutions are possible nor can we construct global shape function based on these approximation, so these are some of the difficulties. Moreover we would like to deal with certain complexities in our studies which could include one or more of this, for example we would like to handle geometric complexities, curved shapes, cutouts, large size structures for example nuclear power plants, cooling towers, bridges, long span bridges and so on and so forth with foundations and piers and superstructure all of them treated together. So geometric complexity is a basic difficulty that we wish to handle, similarly in homogeneity and structural material and geometric properties, say members need not be prismatic, thickness could vary, material properties could vary and so on and so forth. Then we may even deal with structures assembled out of simple structural elements like bars, beams, plates, shells and 3D continue, for example a bridge structure will have soil, piers, superstructure and so on and so forth, so different parts of the structure could be modeled using different structural elements, so but we want to treat the entire assembly as a single structure. Then we want to deal with diversity of structural behavior, for example static stress analysis, vibration analysis, buckling, fracture, fatigue, creep, parametric instabilities, bifurcations and so on and so forth, so when we deal with practical problems we should have handle on dealing with some of these issues. Another important issue is we would like to be able to deal with geometric or material nonlinearities and time varying systems, we will come to that examples of all this in due course. There are many problems where more than one physical process is involved, for example in fluid structure interactions like pipes conveying fluids or liquid storage tanks subjected to earthquakes on and so forth, we would like to model the interaction between structure and the fluid. In thermal stress analysis which is needed for example in analysis of structures under fire, we need to combine heat transfer analysis with stress analysis and in electromechanical systems like smart systems and things like that the field variables could include certain electrical variables in addition to mechanical variables. Similarly structures under active control, for example in buildings where we have active control mechanisms in place against earthquake loads. So how do we deal with this variety of problems? We need a general framework where we can formulate these problems, analyze and try to interpret the results. Now the finite element method is basically a powerful means to tackle some of these issues. The main idea of finite element method is that the field variable defined over a given domain is approximated in terms of piecewise polynomials over subdomains and we assemble the solutions that we develop on subdomains suitably to get the solution for the entire domain. So the use of piecewise polynomials lie at the heart of finite element method. Now the finite element method has developed significantly in recent years, the main drivers for this is advances in computer hardware and software and we have now professional softwares for using finite element method as a tool which has three important components. It is not only that we use computers to solve the governing equations but we also use the computer as to even to formulate the problem, set up the mathematical model for a given structure by taking into account geometric details, constitutive laws and so on and so forth and then arrive at the governing equations that need to be solved. The equations that need to be solved could be a set of algebraic equations or a set of coupled ordinary differential equations. It is not that we use computers only to solve those equations, similarly after we solve these equations we use again the computers to process the results. So it is not that we obtain solutions at all points on the structure in our analysis but we need to know what could be the response at different points of the structure and how to display the results so that engineering decisions can be made. So computers are being used extensively to deal with all these wide variety of problems and this contributes to the wide success of the method. On the other hand the mathematical foundations of the subject is also being well understood and now study of finite element analysis forms a major theme in mathematics research. So let me ask this question, finite element method we have abbreviation FEM what does this, what do these three alphabets mean? What is F in finite element method that is what is finite in finite element method, what is element in a finite element method and what is this method in the first place? So we will address this problem of what is M, finite element method is an approximate numerical method for analyzing mathematical models of physical phenomena encapsulated through a set of partial differential equations or variational principle. So it is an approximate numerical method basically to treat partial differential equations or problems that can be formulated using certain variational principle. Now we also have the phrase finite element analysis, now this can be taken to mean study of mathematical aspects of the method including questions on how the various approximations may affect the convergence and accuracy of the solutions. So this word analysis implies these addressing these questions whereas method implies using the tools to solve engineering problems with a proper understanding of how the method has been developed in the first place. Now to explain what is finite in finite element method we will start with some observations, we generally work in Cartesian coordinate system when we apply finite element method we typically deal with vectors and tensors like displacements and stresses and strains and so on etc. Therefore we need a coordinate system, so we use Cartesian coordinate system. Now a field variable U of X tilde, T let us say that it is a S cross 1 vector of dependent field variables which we need to determine and omega is the domain of interest. So X tilde is X1, X2, X3 which is a Cartesian space. Now what finite element method does is it treats we don't take the field variable U X tilde comma T as unknown, what we do is we select a certain points XI tilde and the value of the field variables at these points if there are N number of them these are treated as unknowns and we aim to determine this and after knowing this we will develop an approximation to the field variables in terms of these N variables. So the method employs and also to develop these solutions I mean for this purpose the function method employs piecewise polynomials to achieve that that is again a very important aspect of the method. So therefore the problem of approximating the infinite number of unknowns that is U of X tilde comma T for all X tilde belonging to omega and for all T greater than or equal to 0 is now converted into the problem of determining a finite set of functions U of XI tilde comma T I running from 1 to N, so this is what makes the and the name finite in finite element method, this is what this aspect of the solution contributes to this nomenclature. Now the points in the domain where we the field variables are taken as unknowns are known as nodes, now we also call this as the value of the field variable at these points are degrees of freedom at the node I, these are the generalized coordinates which have typically have a direct physical meaning because they are the values of the field variable at the nodes they have immediate physical meaning because the field variable itself has a physical meaning for in a given problem. Now in this discussion the time variable T is still continuous at this stage, I have not talked about discretizing time so we call this discretization as semi discretization only spatial variables are discretized the time is still treated as a continuous variable. So now the question is how this U of X tilde comma T is approximated in terms of these degrees of freedom, this leads to the notion of element that the E in the finite element method, what does that mean? So you consider a domain omega, suppose I partition this omega into a set of I equal to 1 to R as shown here, this is one Ith element, we say that these omega I's whose union contributes to the domain and we expect that there is no intersection between these sets, that means omega I intersection omega G is a null set for all I not equal to J and omega I is known as the Ith element. So this element itself now need to be characterized in terms of geometry and how many nodes it has, I have shown for each element here for this type of element 3 nodes but whereas if you look at this there are 3 nodes, then how many field variables are treated as unknowns at a given node that depends on problems as we will shortly see, and what is the total number of unknown coordinates that is assigned to each element that is known as total degrees of freedom for the element, then value of the field variables within the element, how do we determine? Suppose I know value of the field variable at the nodes, how do I get the field, value of the field variable within an element, so these are some of the issues that we need to address when we talk about an element, so what we do is suppose if you consider this element I name these 4 points as nodes 1, 2, 3, 4, and what we do is within an element we approximate the field variable by interpolating the value of the field variables at these 4 points. So we write this as J equal to 1 to Q, Q is 4 in this illustration and this is the value of the field variable at 1, 2, 3, 4 and these are known as the interpolation functions, so NJ of X tilde are interpolation functions, what we do is we choose these interpolation functions to be polynomials, the nature of this choice is paramount in determining the quality of the solutions we get, so this is a very vital aspect of implementing the method, so what are the typical steps in finite element analysis of say structural dynamic problem, so what we do is partition the domain into a set of elements, now this domain omega actually we may not be able to represent exactly in a finite element model, for example it has curved edges so in our modeling even the domain is replaced by an approximation, so those details will come to as and when we encounter curved boundaries and things like that, so even these curved edges need to be suitably approximated, it is not that the given domain omega with all its complexity is exactly replicated in our model, so that also should be emphasized, so that we partition the domain into a set of elements, then we approximate the behavior of the field variable within each element either starting from the governing partial differential equation or the relevant variational principle if it is available for the given problem. We use trial functions, as I said trial functions are piecewise polynomials, this leads to element level structural matrices, typically in dynamics we get element mass stiffness and damping matrices and vector of nodal forces, next we have now partitioned the domain and analyze each bit like this, now the next job is you should put back all this and form the built up structure, so we assemble element level matrices and force vectors to form the global matrices and force vector for the structure, this is assembling, then we look at the boundary conditions to be imposed, right, after we impose the boundary conditions we obtain the governing set of equilibrium equations for the problem, typically in this application there will be a set of ordinary coupled, second order coupled ordinary differential equation, they could be non-linear if the underlying problem is non-linear otherwise it could be linear, now for these set of ordinary differential equations we now address the question of discretizing the time, so we typically use time-stopping methods explicit or implicit depending on the details of the problem and this completes the determination of the field variables at the nodes, then we compute derived quantities like stresses and strength, so displacements are the degrees of freedom which in most of the applications is the case, then we need to derive strains, stresses at the nodal values, at the nodal node points as well as in the interior, okay, so that completes the solution. We need to consider several other points, I will just make a brief mention of these points and we will return to some of these issues as we go along in the course, so we have said that we have defined element through this description, so the given domain is partitioned into a set of R subdomains which are mutually exclusive and collectively exhaustive and we call omega as ith element, this also defines the division of geometry, now the word mesh is used to denote the collection of elements which replaces a domain omega in the simulation of the node, now we need to address question on inter-element behavior of field variables, so we should note that the communication between two distinct element is only through the value of the field variables at the nodes, it is only through the points, common shared points, we don't talk about interaction through lines or surfaces, it is only through points even in two dimensional and three dimensional problems, because in two dimensional, in a line element the two elements are connected through a point, but in a plane element, two dimensional element the two neighboring elements can be connected through lines, and in solid element, three dimensional element the neighboring elements are connected through surfaces, but the interaction is always through nodes that are the points. Then how do we evaluate the derived quantities, for example strains and stresses within an element and how do these quantities behave across the element boundaries, so this is an interesting question, we will see that there are some difficulties in modeling this and this is one of the concerns that we should be aware of, then we should look at quality of the answers obtained, so there are many decisions that we have made, how many elements that we should take, that means how fine should this partition be, that we call it as H, H indicates in some sense the size of the element, and within an element what order of polynomial we should use to interpolate the field variables, that determines P, and delta T is the time step that we need to use in integrating the semi discretized equations, so the accuracy of the solution finally depends on a combination of H, P, and delta T, so we need to understand how to make these choices to ensure that we get acceptable results. So now to develop this discussion on finite element method the approach that we will take is we will start by developing structural matrices for different types of elements, like actually vibrating bar, torque element, 2 dimensional and 3 dimensional beams, then plane stress and plane strain elements, plate bending element, shell element, 3D solid element and so on and so forth, so for each of this we will develop, we will characterize the behavior of a single element, and then we will address question of how to assemble the formulations at element level, and study structures such as trusses, planar frames, grids and 3D frames, and also certain 2 dimensional and 3 dimensional continuum structures. So this is the broad outline of at least the next few lectures, so we will start somewhere, we will start with addressing questions listed here, as a prelude to that we will quickly review some notations and equations from linear elasticity for isotropic material. So the independent variables and the coordinates of the system we use X, Y, Z or X1, X2, X3 and time T, so there are 4 independent variables in a 3 dimensional problem of elasticity, one is time and 3 are spatial coordinates. Now we can make a coordinate transformation, so we can have one coordinate X1, X2, X3, a coordinate system, another one X1 prime, X2 prime, X3 prime, and we would like to know how to transform quantities described in X1, X2, X3 to the new coordinate X1 prime, X2 prime and X3 prime. The coordinate transformation matrix for a position vector is given by this well-known transformation matrix, this is an orthogonal matrix we will be using some of this quite often. What are the dependent variables in elasticity problem? The state of stress at a point is given by the stress matrix which is 3 by 3 symmetric matrix, and each component will be function of X1, X2, X3 and time T. Then we also have strain components which is again a tensorial quantity and it is described by another 3 by 3 matrix which is again symmetric and which is again function of the 4 independent variables. Displacement component, there are 3 displacement components U1, U2, U3, and they are again functions of the spatial and time coordinates. So the total number of dependent variables is 15, independent variables are 4. Now how do these dependent variables behave under coordinate transformation? Sigma stress and strain are tensorial quantities, so they undergo transformation through these rules, where T is the transformation matrix between the coordinate system. So sigma prime is the state of stress at a point in the transform coordinate, new coordinate system, and this is the sigma is the state of stress with respect to a reference coordinate system, similarly strain, and this is displacement, displacement is a vector, this is how it translates, scalar quantities are invariant with respect to coordinate transformation. So now we have 15 unknowns and we need to now generate the 15 equations to tackle this problem, as we know there are 3 equilibrium equations, 6 strain displacement relations and 6 strain stress strain relations, so I have listed these equations here for a quick reference, the equilibrium equations have terms involving gradients of stresses, and also this capital X1, X2, X3 are the body forces, there are forces which are proportional to the volume, and rho is the density U1 double dot, U2 double dot, these terms on the right hand side represent the inertial forces, okay. So sigma IJ is same as sigma JI, therefore there will be 3 equilibrium equations, the strain displacement relations are given through this set of formulae, again strain epsilon IJ is epsilon JI, these are normal strain, these are shear strains, the 3 normal strains and 3 shear strains. Now for a Huckeyn material the stress and strain are linearly related, so we can show that there will be 2 independent constant, in terms of Lemme's constants the constitutive laws can be given like this, the way the stresses are related to the strains is through this, the E is the sum of the normal strains, and in terms of that we have this constitutive law, these are the stress strain relations, in terms of, you can also express strains in terms of stresses, this is the first invariant of the stress matrix, and we get these relations, again this is in terms of Young's modulus and shear modulus, so we could use either this or this depending on the context in which we formulate the problem. If you use indicial notations all these equations can be written in a compact manner, so this is the equilibrium equation, set of equilibrium equations, this is the set of strain displacement relations, and this is the stress strain relation, IJKL etcetera take values from 1, 2, and 3, so the repeated indices imply summations, a comma is a differential, all those notations are used here. Now there is yet another way of representing stresses and strains, here we write the state of stress at a point as a 6 cross 1 vector, instead of a 3 by 3 matrix, a symmetric matrix we write it as 6 cross 1 vector, and the strain is written in this, so this is not a tensorial quantity anymore, so this is the strain, and if we now write the stress strain relations in this notation, this C will be a 6 by 6 matrix, because stress is a 6 cross 1 vector and epsilon is 6 cross 1 vector, so either we can write stress in terms of strain or strain in terms of stress, and these matrix obey this identity and we have these expressions for the C matrix and C tilde matrix, and nu is the Poisson's ratio is a Young's modulus, so this is for an isotropic material. Now if we define an operator D as shown here, this is 6 cross 3 matrix operator, we can write the governing equilibrium equations and in this form D transpose sigma plus X is rho U double dot, now the strain displacement relation using the same matrix can be written as D U equal to epsilon, now here if I now use the consultative law sigma is equal to C epsilon or epsilon is C tilde sigma, and suppose if I substitute here, so I will write for sigma C E then this equation become D transpose C epsilon plus X equal to rho U double dot, now for epsilon I will write, for epsilon I will write now D U, so if I write that I get this as the equations, governing equations where the equilibrium equations, stress strain relations and constitutive laws are all been captured, and this the final equations are written in terms of displacement components, okay, so D is this operator, C is the constitutive matrix of constitutive laws, and this is the body force, vector of body forces. Now we will also need a general approach to derive the strain energy in a body under a multi-axial state of stress, so we will start by discussing how to derive strain in a uniaxial state of stress, suppose at a point in the structure we choose an infinitesimal element DX, DY, DZ, or volume DX, DY, DZ, and if sigma XX is the only non-zero stress component, the forces on this phase is given by sigma XX DY, DZ. Now the work done by these forces on the deformation can be computed over this entire volume, the force is sigma XX DY, DZ, and the displacement is strain into DX, this is this, and half because the load is applied gradually, and we can write this as sigma XX, epsilon XX, DV, integral over V multiplied by half. Now if the body is in a multiple state of stress, that means all stress components are acting simultaneously, then this logic can be extended and the equation for strain energy over the entire volume can be written in terms of the stress vector and strain vector in this form, okay. So now if we substitute the constitutive law and use the strain displacement relation, the strain energy can be expressed in terms of the displacement components, the operator D and the constitutive matrix of constitutive coefficients in constitutive law, so this is an expression for strain energy in a 3D solid under multi state of stress, so this is strain energy, this is the kinetic energy, right, so the Lagrangian is this and we can define the action integral, so if we now apply the, you know select U so that action integral is minimized, now we recover the equations of elasticity, okay. So this is a general framework for treating 3 dimensional elasticity problems, now as a special case we will start now talking about simple elements, we will start with discussion on axially vibrating rod element, so I consider a rod element of length L, axial rigidity AE damping C and M is mass per unit length, so this is a cross section and the coordinate system X is along the length of the bar, Y is as shown and Z is shown here. Now this rod will be in a state of uniaxial state of stress, therefore the state of stress is given by sigma XXX, T that means all other stress components are 0, so this is a standard assumption that we make in modeling axially vibrating rods, so strain is consequently given by this and the displacement is only function of X and T, now if I use that now in the expression for strain energy we will get now the strain energy expression as this kinetic energy is this and the action integral is this. Now we want to now develop the finite element approximation to handle this element, so what we do is we define these two ends as nodes, so that I have shown as these numbers in a circle 1 and 2 are nodes, U1 of T and U2 of T are the values of the field variable U of X, T at node 1 and node 2 respectively, P1 of T and P2 of T are the stress resultants at 1 and 2. Now we will consider the equilibrium of the bar under the action of support displacements U1 and U2, so suppose phi 1 is the trial function that we are going to you know work with, we need phi 1 of 0 that means what we are doing is we are going to approximate the field variable in terms of the value of the field variable at the nodes that means we are interpolating suppose I want a state of stress here, the field variable here I will interpolate the values at these two nodes and this is a representation, and phi 1 of X and phi 2 of X are the trial functions. Now how to select these trial functions that is a question we are asking and what we do is we will assume phi 1 of 0 is 1 and phi 1 of L is 0, this is the equation under these displacements and we get phi 1 of X is 1 minus X by L, similarly we get for the other displacements phi 2 of 0 is 0, phi 2 of L is 1 we get phi 2 of X as X by L that means we are basically using linear interpolation functions, so what we are doing is we are taking U of X, T as U1 of T into phi 1 of X plus U2 of T into phi 2 of X, U1 of T and U2 of T are the generalized coordinates, we call phi 1 of X and phi 2 of X as trial functions, so now U of 0, T is U1 of T because at X equal to 0 the prescribed displacement is U1 of T, so if we use that U of 0, T is U1 of T, phi 1 of 0 plus U2 of T, phi 2 of 0, so consequently I get phi 1 of 0 is 1, phi 2 of 0 is 0. Next U of L, T is U2 of T, so if I impose that condition I will get phi 1 of L is 0, phi 2 of L is 0, now this ensures that if we select now phi 1 and phi 2 so that these 4 conditions are satisfied, this ensures that we have satisfied the requirements on displacements at X equal to 0 and X equal to L, now the previous slide I discussed on how to select these phi 1 and phi 2 under the required conditions that is this one and this one, so now we are ready with the interpolation functions which are 1 minus X by L and X by L and then approximation for the field variable as shown here, now I can find the velocity field by differentiating this with respect to the time I will get U1 dot of T into this U2 dot of T into phi 2 of X, similarly I can get the derivative with respect to X, this will be this. Now I can substitute this into my expression for strain energy and kinetic energy, the strain energy is given by half AE dou U by dou X whole square, so if I now substitute that dou U by dou X as you have seen here it is U2 minus U1 by L it is independent of X, so I can quickly evaluate this, next kinetic energy this is half M U dot square, so U dot is U1 of T 1 minus X by L U2 of T X by L whole square DX, now these are polynomials simple polynomial so I can quickly integrate this, so if I do this I get this as the kinetic energy in terms of the two generalized coordinates and this is the strain energy in terms of the generalized coordinates, so now I can form the Lagrangian, the Lagrangian is T minus V I get this, there are two degrees of freedom and I need to run the Lagrange equation for U1 and U2, so that involves finding gradient with respect to U1 dot and U2 dot and substituting these terms we can easily derive all this and we get the equation governing U1 to be this, equation governing U2 to be this, and now I can recast this in a matrix form I can write an acceleration vector U1 double dot plus U1 double dot and U2 double dot, and this gets multiplied by the so called mass matrix and U1 and U2 which is a displacement vector which gets multiplied by the stiffness matrix, so I will come to this force vector in due course but we can believe that these are the nodal forces, so I get now the equation, element level equation of motion of the form MU double dot plus KU equal to P, now we can make few observations, this M which is ML divided by 6 into this is known as element mass matrix, we also call this as the consistent mass matrix, why? What is the genesis of the phrase consistent here? It is consistent mass matrix because the shape function that we have used for computing the kinetic energy is same as the one used in computing strain energy, so that means in that sense it is consistent, it also implies that there is an alternative way of finding mass matrix which is not consistent I will come to that shortly, and you can see that mass matrix is symmetric, and this quantity K AE by L into this matrix is an element stiffness matrix, it is symmetric, and we call this element that we have developed as a 2-noded bar element with 1 degree of freedom per node and 2 degree of freedom for the element, this K and M we should also notice that they are non-diagonal, they are non-diagonal matrices, so the nodal degrees of freedom are coupled in this representation. Now I talked about consistent mass matrix, an alternative way of deriving mass matrix used to use heuristic arguments, and in an element half the mass is lumped at one end and half the mass is lumped at the other end, if we do that we get mass matrix to be diagonal and this is called lumped mass matrix, this is diagonal but it excludes inertial coupling between the nodal displacements, the earlier formulation showed that these quantities are not 0, so obviously that coupling has not been considered in this formulation. Lumping is to some extent an arbitrary process, in this case it may seem intuitively acceptable but when it comes to elements where there will be flexure as we will soon see, lumping this mass matrix may not be straightforward. Now we can make some more observations, we have now this representation U of X, T is 1-X by L U1 of T plus X by L U2 of T, suppose both the nodes are given equal displacement of unity, that is the rod is undergoing a rigid body displacement, in that case what should be U of X, T? It should be 1, because there should be no elastic deformation of the bar and that is consistent, we do get that, so this means phi 1 of X plus phi 2 of X is 1 and we call this property as partition of unity. Now let's again take a look at the assumed representation for the field variable, now let us, this is a displacement let me compute the strain, we get strain to be independent of X, okay, so strain is independent of X for any L, so that means constant strain condition can be simulated as element length goes to 0, so in a finite element model as element becomes smaller we expect that there won't be any strain variation within an element and that is simulated correctly here, so these two properties are of, they have serious implications on convergence of finite element solutions and at some later stage we will revisit these issues again, but at this stage we can make these observations. Now we are ready to tackle some quick examples, so suppose there is a rod which is clamped at this end and it carries an axial load like this, so suppose I model this using one element we see that this bar is clamped here therefore U 1 of T is 0, and at this end I am applying an external force P of T therefore P 2 of T should be P of T, now if I put that into my governing equation for the element I get for U 1, U 1 double dot is 0, U 1 is 0, and at P 1 that means at this end there will be an unknown reaction, okay, so that is R of T which is not known, on the other hand there is an applied force that remains as P of T and U 2 is the field variable U 2 of T is not known, so there are essentially two equations for two unknowns, one equation is for the unknown nodal displacement U 2 of T, other equation is for the unknown reaction at node 1, so now if you write the second of these equations I get the governing equation for the unknown nodal displacement and, sorry this is a equation for reaction the first equation I get reaction in terms of U 2, now the equation for U 2 itself can be obtained by considering the second row here and we get this equation, so if you carefully look at the structure of these equations it follows that first you need to find U 2 of T using this equation, and then substitute back into this to get the reaction, so if you look at this equation now I get this is a single degree of freedom approximation there is only one generalized coordinate, so I get this an approximation to the natural frequency according to this model is this, and U of X, T is X by L U 2 of T, U 2 of T is solution to this, and the strain is given by this the stress is given by, the axial thrust is given by, so we have found out now displacement, strain, stress, stress resultant and the reaction, so this framework enables you to do that. Now let's consider the situation to understand the nature of approximation that we are making it's useful to consider a bar as shown here, and suppose I make 5 divisions here, see the statement I made was in finite element method we use piecewise polynomials to approximate the behavior of the field variable for the entire structure, so let us make now 4 subdomains as shown here, and within an, the U of X, T is the unknown displacement field all over this for X varying from 0 to L, and this is written like this, so when viewed like this what are the shape function, what are the trial functions, the first trial function in our model will be this, this has nonzero value only over for the first element and it is 0 here, it is linearly varying within the, the second element, the second function psi 2 of X is in terms of these 3 nodal coordinates and this is linearly varying as shown here, so similarly the other elements are shown here. So what we are done, what we have done just now is we have considered the behavior of only one element and got the equation of motion for this element, but we need to actually study a built up structure like this, so we need to find equations for A1, A2, A3, A5, now we'll come to that in due course but I want you to make note of some simple things, see what we are doing is that the element level we are using these as the trial functions, but at the built up structure level these are the trial functions, but they are all related. Now how do we see that suppose I take the element E the way I am interpolating UE of T is through these 5 functions it is in terms of the element nodal coordinates and the element level trial functions, so this is the kinetic energy for the element and this is the potential energy for the element, so the superscripts E here refers to the element E, now that means for each element I can do this exercise of computing the strain energy and kinetic energy using the local interpolation functions, and when it comes to computing the total strain energy and the kinetic energy I can add contribution to this from each of these elements as shown here, okay, so this is what we will be doing when it comes to question of assembling the matrices. Now we can extend this logic now to problem of Euler Bernoulli beam, so I will begin discussing this and we'll continue with this in the next class, so the problem description is as shown here this is one beam element, it's characterized in terms of flexural rigidity EI mass per unit length M, length L and C is the damping properties and this is the X direction in which the displacement field is U, U of X, T and this is the Y direction where displacement field is V, now we are therefore considering bending in XY plane we are making an assumption that the X axis coincides with the centroidal axis of the beam, and we also assume that there is no coupling between bending and torsion, this cross section is symmetric therefore there is no coupling between bending and torsion. So if we now follow the tenets of Euler Bernoulli beam hypothesis U of XY, T is given by this because plane sections which are normal to the neutral axis before bending remain plane and normal to the neutral axis after bending, so if you make those assumptions if you use those, if you use that assumption we can write displacement like this. Now V is a function of X, T, now from this I can get strains, this is the normal strain, this is the shear strain, shear strain turns out to be 0 in this model, so I can compute the strain energy, we need to use sigma XX and epsilon XX and if you use this is an integral over volume, so now I will integrate over X and area of cross section separately, so using the assumptions that we have made we get for epsilon XX whole square minus Y dou square V by dou X square whole square, and that remains here, and the remaining integral that is integral AY square DA is nothing but the moment of inertia about the Z axis so I write this equation, so this is a strain energy stored in the beam due to bending. The kinetic energy is integral over volume rho V dot square DV, and again the DV if I write it as DX into DA I can write M as integral A rho DA this will be the mass per unit length so I will get the familiar equation for the kinetic energy, so the Lagrangian for the system is T minus V and this is the expression for the Lagrangian, so in the next class what we will do is we will consider the beam element to have two nodes, and at each node we will consider field variables to be the translation and rotation, this double arrow notation is for rotations, so this is a rotation using right hand rule convention, so U 2 and U 4 are rotations, U 1 and U 3 are translations, so collectively we call them as displacements, so associated with them there are stress resultants, so we assume the field variable which is V of X, T which is a transverse displacement of the beam in terms of the nodal translations and rotations, and we need to therefore construct a solution, we need to construct four interpolation functions to achieve this, so we will see how this we can do in the next lecture.


This talk is part of a series giving an introduction to complex projective surfaces and this one will be mostly about minimal surfaces. This is minimal surfaces in algebraic geometry. They have almost nothing to do with minimal surfaces in differential geometry. So we recall that every algebraic surface is birational to one that is projective and non-singular and minimal. And the classification of surfaces tends to be a classification of projective non-singular minimal surfaces. So I'll just briefly recall how you get each of these properties. First of all, for projective, this is easy. All you have to do is take an open affine subset, s naught of your surface which is birational to the whole surface, and then s naught is almost by definition contained in some affine space which is contained in some projective space. And you can take the closure of s naught in some projective space and this will give you a projective surface. This projective surface might have singularities, even if s naught doesn't have singularities. So we need to remove the singularities. So the second step is to make it non-singular. Well, making a surface non-singular is actually completely trivial. All you can do is discard the singularities, but of course that will no longer be projective. So the problem is to resolve the singularities while still making sure the surface is projective. This is a really hard problem called resolution of singularities and it was solved by Walker for characteristic zero and by Zariski for characteristic greater than naught. There were several claims of having done this before Walker, but none of them seemed to have been quite complete. So they did this for surfaces. Resolution of singularities and high dimensions is an extremely difficult problem that there's been a lot of work on. So Hironaka didn't characteristic zero and positive characteristic is still open. So I just quickly sketch how you resolve singularities and characteristic singularities of surfaces. What you can do is we can remove co-dimension one singularities by normalization. So normalization is a operation you can do which on affine schemes just corresponds to taking the integral closure of an integral domain in its quotient field and you can do this for all schemes and it turns out to be slightly stronger than resolving all co-dimension one singularities. So you can get rid of the co-dimension one singularities and now you can get rid of the co-dimension two singularities. So let's remove co-dimension two singularities. Well, co-dimension two singularities for surfaces are just points and we can just blow up the points. So we seem to have got rid of all singularities, right? Well, not quite because when you blow up a point, this can introduce new co-dimension one singularities. So you first do step A, then you do step B to get rid of the co-dimension two singularities. Well, it doesn't get rid of them, you blow them up. Then you've got some new co-dimension one singularities. Well, that's okay. You go back to step A and then back to step B and you keep on going and the problem of resolution of singularities is to show this process. So it's fairly easy to write down a method for resolving singularities. The problem is just to prove that it works. For example, you could try to measure the nastiness of a singularity by some suitable invariance and try and show that this process reduces the maximal nastiness of all singularities of the surface and show that you can't decrease the nastiness an infinite number of times. The trouble with this is it's rather difficult to find a good measure of just how nasty a singularity is. In fact, this is almost the essential problem in resolution of singularities. Anyway, we now come to the third step, which is what I really want to discuss today, which is to show that the surface is minimal, which means it's not the blow-up of some non-singular surface. It's quite often the blow-up of the singular surface. So let's just recall what the blow-up is. Well, let's just give the blow-up of the point, the origin of two-dimensional projective space for simplicity. So two-dimensional projective space is set of points with projective coordinates x, y, z. And for its blow-up, we first take the product of this with p1. So p1 has coordinates x, column y. I'm going to use capital letters here. And we take the subset of p2 times p1 of all points, such that x times y equals y times x. And this gives us the blow-up of p2 at a point. And the blow-up has a natural projection map to p2. And there's an isomorphism, except over the point 001. And the inverse image of 001 is, in fact, a copy of a projective line. So we've replaced a single point of p2 by an entire line. What happens is something like this. So here we've got a copy of p2, which I'm going to draw as a plane. And here's the point we blow up. And blowing it up kind of changes it like this. So we replace the point by an entire copy of p1, which I'll draw as a little circle, though p1 isn't really a circle, but I can't think of what else to draw it as. And what is going on is that each line through the origin corresponds to a single point of this p1. I mean, the way I've drawn it looks like it corresponds to two points, but that's because I'm not very good at drawing in four dimensions. So what we have here is a bi-rational map, sorry, is a regular map from the blow-up of p2 at a point to p2. And this map is actually bi-rational. So that's what blowing up a point is. And we want to show, find a minimal surface, which is one you can't obtain by blowing something up. Well, the obvious thing to do is to start with a non-minimal surface and blow down copies of p1, which is an obvious inverse of this operation, and keep blowing down points until we can't do so anymore. And the problem is, is this process finite? And the answer is it is finite. And one way of seeing this is that blow-up at a point increases the dimension of the second homology group of a surface by one. It's not too difficult to see that because blowing up a point is replacing the point by a copy of p2, sorry, a copy of p1. And the projective space p1 over the complex numbers has a second co-homology group of rank one, which is where this one comes from. So blowing down decreases the dimension of the second homology group by one. And since the rank of the second homology group is finite, you can't blow down more than a finite number of times before you come to a stop. So this shows that every non-singular projective surface is bi-rational to a minimal non-singular projective surface. And it's the minimal ones we really try to classify. And now I'll say a bit more about what the structure of bi-rational maps is. So here we want to talk about the structure of bi-rational maps. This was worked out by Zariski, although it was sort of implicitly known to the Italian algebraic geometers. I think they just didn't state it explicitly. So I suppose we've got a bi-rational map from the surface s0 to s. So we might take s0 being p2, for example, and s1 being p1 times p1. And these two surfaces are bi-rational but not bi-regular. So what Zariski said is that you can turn this into a regular map by blowing up s0 repeatedly. So we blow up a finite number of points. And these need not be points in s0. For instance, s1 might you might blow up a point in s0, but then s2 you might blow up a point in s1. And this might not be a point in s0. It might be a point in the in the extra copies of the projective line you get. So he said you can do that a finite number of times. So this map is now regular. So that factorizes any bi-rational map into a sort of inverse of a lot of blow-ups at points followed by a regular map. This isn't too difficult to prove. The idea is that if you've got a bi-rational map, it's sort of defined everywhere except at some points. And you can make it more defined by blowing up these points. And if you keep careful track of this, you'll find it can be factored like that. More generally, any bi-rational map can be written entirely in terms of blow-ups. So we can, in fact, write if we've got a bi-rational map from s0 to s. So this is bi-rational. We can find a sequence of blow-ups and then a sequence of blow-downs. Let's call this tn minus 1 down to s equals t0. Such that the bi-rational map is got by first of all blowing up a number of points and then blowing down several times. So in some sense, if you can understand blowing up at a point, then you've sort of understood every bi-rational map. So now let's look at this operation of blowing down. So what we want to do is ask, when can you blow down a surface? So we've asked, when can we blow down surface s? In other words, we want to give an s, can we find a map from s to s0 such that s is the same as blowing up s at a point? And so this is blow-up at, say, a point p in s0. And we call the inverse image of p. Let's call it e is called the exceptional curve, or the exceptional curve of the first kind, if you want to be really pedantic, but we won't worry about exceptional curves of the second kind. So when can, we can ask the question, when can we blow down a curve e? And the answer was given by Castelnuovo. He said, well, there's one obvious property. First of all, e must be rational. e is isomorphic to a projective line. It's kind of obvious. The second is a much more subtle property, that e equals minus one, where this is the self-intersection number. And the easy part of the theorem is to show that if you blow up a point, then the exceptional curve has self-intersection minus one. The hard part of Castelnuovo's theorem is to show that conversely, if you've got a curve like this, then you can blow it down. So what I need to do now is to show that, is to show what the intersection number of two curves is. So let's talk about intersection numbers of curves. So we have the following problem. Define the intersection number cd of curves, cd on a surface s, where s will be projective and non-singular. And I'm going to describe three ways of doing this, all of which have some disadvantages. The first is the naive definition, where we just define cd to be the sum of the, well, let's define it to be the number of intersection points. Well, there are several problems with this. So first of all, if the curves look like this, there's absolutely no problem that there's a single intersection point. On the other hand, if the curves look like this, then we kind of have to count that as a double intersection point. So this shouldn't quite be the number of intersection points. It should be the sum of intersection points with some sort of multiplicity. And the multiplicity can get quite complicated. For example, one of the curves might have a singularity. Well, obviously then you calculate two intersection points, but what happens if both curves have a singularity and they meet at the singularity? What on earth are you going to count this multiplicity as? And it's not entirely obvious. But there's an even worse problem. What happens if you take the intersection of this curve with this curve? In other words, if you take the intersection of a curve with itself, there are an infinite number of intersection points. So what are you going to do? Well, what we can do is deform one curve. So we now take the intersection multiplicity of this curve with a slightly different curve. We might sort of shift it very slightly, that might end up like that. Well, can we do that? Well, there's one problem. Can we deform the curve? And the second problem is the intersection multiplicity independent of the deformation. So you see we're actually getting quite a lot of problems in trying to define the intersection multiplicity. The answer to question one, can we deform the curve, turns out to be not always. So already we've run into yet another problem that we have to think of something else to do. If we can deform it, then you can at least show that this is independent of the deformation. So although you can push this through, it's really rather difficult and it's less effort to use a different, less intuitive definition. So here are some easier definitions. First of all, you can map any curve to a cohomology class in the second cohomology of your surface S with real coefficients. And the second cohomology class has a cut product from the second cohomology group times the second cohomology group to the fourth cohomology group, which is isomorphic to the reals. So this gives you some sort of bilinear map on curves. And this does give you a reasonable intersection multiplicity. One obvious problem is it involves working with cohomology over with real cohomology, which is rather hard to make sense out of over fields of characteristic greater than zero. You can do it using etal cohomology, but it's a lot of work. Fortunately, there's a more algebraic definition that's easier to work with, where you define the intersection multiplicity of two curves on the surface. So here's C and D are curves on the surface, and we define it to be the Euler characteristic of O minus the Euler characteristic of O of minus C minus the Euler characteristic of O of minus D plus the Euler characteristic of O of minus C tensor with O of minus D. So what on earth does this mean? Well, this is the sheaf of regular functions on the curve. This is a line bundle associated to a divisor, and this is the tensor product of line bundles. And chi of a line bundle or invertible sheaf is the Euler characteristic, which is the dimension of the zeroth cohomology group minus the dimension of the first cohomology group plus the dimension of the second cohomology group. This is coherent sheaf cohomology, which is in chapter three of Hartron's book, and which I strictly speaking, we haven't yet covered in the course. Well, this definition actually works great. There's one obvious problem with it. The main problem with it is that looks weird. It's completely unintuitive and not at all clear how you would ever think of this. The advantage of it is it's actually really easy to work with, because cohomology groups, although they're a bit unintuitive, are actually very easy to work with, and the Euler characteristic is very nice properties. So if you're willing to put up with the fact this is not at all intuitive, this is actually the easiest definition of the intersection multiplicity, and then you can then go off and prove a theorem saying that it's the same as the sum of the multiplicities of the intersection points and so on. So next I want to discuss how is it possible for a curve to have negative self-intersection multiplicity? So how come we can have cc being negative? There's certainly no problem having it equal to zero. If you look at p1 cross p1, we see we can think of this as a lot of horizontal lines times a lot of vertical lines, and the any two vertical lines obviously of intersection multiplicity zero and any two horizontal lines of intersection multiplicity zero. So intersection multiplicity zero is no problem, but what about less than zero? Well, let's prove that we can't have two curves with intersection, a curve with self-intersection multiplicity less than zero. So suppose we've got a curve c. What we do is we deform c slightly to get a curve c prime, and now intersection multiplicities is invariant under nice deformations of curves. So we see now that cc is equal to cc prime, which is greater than or equal to zero because it's just a sum of intersection multiplicities of a finite number of points. So this seems to prove that curves cannot have negative self multiplicity. Well, there's a problem here. What this actually proves is that if cc is less than zero, we cannot deform c to have finite intersection with with itself. In other words, for instance, c might just be rigid. There's no way to deform it, or c might have two components, and you can deform one of these components, but can't deform the other one or something like that. Well, when I say we can't deform c, what I mean is we can't deform c algebraically. We can deform c topologically. So the difference is we can't deform a curve c to be an algebraic curve inside the surface. But if we think of the surface as being a four-dimensional topological manifold, we can deform c as a topological manifold. Well, doesn't this give us the same contradiction? Well, no, because the topological intersection multiplicity of two surfaces in some four manifold can be less than zero. Because topological intersection multiplicities, you have to sort of worry about orientations and things like that. And intersection multiplicities can be less than zero. For instance, if we're just talking about curves in a, sorry, real one dimensional curves in a real one-dimensional surface, if you've got a curve like this, these have intersection multiplicity one, but if you deform the curve to look like this, it now seems to have intersection multiplicity three, but you have to give this intersection number multiplicity one, and this minus one, and this plus one, and so on. So so algebraic intersection multiplicities of, at a point always finite, but you can't always deform curves, topological intersection multiplicities can be negative and you can deform curves. Anyway, so that's why the intersection multiplicity of a curve with itself can actually be negative. Now I want to show informally why is the intersection multiplicity of an exceptional curve equal to minus one. So remember, exceptional means it's a blow up of a point. Well, I'm going to explain this by look at the case of blow up the origin of two-dimensional projective space. So let's draw a picture of the blow up of two-dimensional projective space. So what you think of this is two-dimensional plane, except the origin has kind of been replaced by a copy of p1. And now I'm going to take two curves two straight lines C and D in p1. And you see that these two lines are generally going to have intersection multiplicity one rather obviously. Now I'm going to move these two lines so that in the plane they both pass through a point. Well in the in the blow up something a bit funny is going to happen because if I move this line D so it passes through the origin then in the blow up of p2 it becomes two lines. First of all we have the entire exceptional curve and secondly we have a line D zero. So what we're really getting is a line D zero plus a copy of the exceptional curve. And similarly if I take C and move it along then it suddenly splits as a copy of some line C zero plus the exceptional curve. So here what I'm doing is I've got a map from the blow up to p2 and I'm moving lines around in p2 and looking at the inverse images in the blow and the intersection multiplicity turns out to be invariant if you move curves around like this. So so C D should be equal to C naught plus E intersected with D naught plus E. And as this should be bilinear it's equal to C naught D naught plus C naught E plus E D naught plus E E. And now we can work out what these are. So C naught D naught is just zero assuming C and D have different tangent directions at the origin because they intersect the exceptional curve in different points so this is equal to zero. And C naught E is obviously just one because C naught intersects the exceptional curve in one point corresponding to its tangent direction at the origin. So this is equal to one and this is equal to one and this sum is equal to one because it's C naught D naught. So if we have one is equal to zero plus one plus one plus this so this must be equal to minus one in order to make everything add up. So that's a sort of informal explanation of why exceptional curves have intersection multiplicity minus one. Now let's look at an example. So let's blow up p2 in two points and see what happens. So here's a copy of two-dimensional projectors space and I'm now going to pick two points p and q. And what I'm going to do is I'm going to look at a line L. So here I've got a line L in p2 and I'm now going to move this line so that it passes through both of these two points. And now let's look at what happens in the blow up. So in the blow up we've got p2 blown up at these two points so let's try and draw a picture of it. So we replace these two points by little copies of p1. So here we've got a copy of p1 and here we've got another copy of p1. And these are the exceptional curves. I'm going to call this exceptional curve e1 and this exceptional curve e2. And now when we move the line L so it passes through both points we see that the inverse image of L in the blow up looks like a line L0 plus e1 plus e2. And now let's work out some intersection multiplicity. So we have one is equal to the intersection multiplicity of L with itself and this should also be equal to L0 plus e1 plus e2 intersect with L0 plus e1 plus e2. And if we work this out it's equal to L0 L0 plus 2 L0 p plus 2 L0 q plus 2 pq plus p sorry that should be e1 that should be e2 and should be plus 2 e1 e2 plus e1 e1 plus e2 e2.  So the conclusion is that this is equal to minus 1 so L0 is rational and it has intersection number minus 1 with itself so L0 is an exceptional curve what this means is we can now blow down along L0. So let's draw a picture of what's happening here so what we've got is we've started with a copy of the two-dimensional projector space we blew it up at one point to get projector space with one point blown up and then we blew it up at a second point to get projector space with two points blown up and this now contains a line L so we now blow down along this line L0 and we ask what do we get here and the answer is what we get here is a copy of p1 cross p1 and let's see why this is true well if we look at p2 and look at these two points we can see through each of these points there's we can draw a sort of p1 worth of lines because all the lines going through that point form a copy of p1 and similarly through this point we can draw lines forming a copy of p1 so we've got a second copy of p1 here and now we can map p2 to the product of these two ones so we get maps to p1 times p1 and this is this map is very easy what we do is if we take any point it lies on one of these lines through the first point and one of these lines through the second point so it's giving us a point from the first p1 and a point from the second p1 so we've got a map from p1 blown up at two points to p1 cross p1 because of course this map isn't quite defined at this point of p1 because it lies in all these lines you have to blow it up to make it well defined well now let's look at the line that goes through these two points and now you see this yellow line this entire yellow line maps to a point in p1 cross p1 because any point on this line lies on the same two lines through the point here so this is of course the line we blow down so this sort of shows informally that if we contract this line to a point then we're probably getting a copy of p1 times p1 so this is an example of a bi-rational morphism from p2 to p1 cross p1 so this is mapping x colon y colon z to x colon 1 y colon 1 in p1 times p1 and of course this isn't always defined it's not defined at all it should be a z so this shows that a bi-rational map from p2 to p1 cross p1 can be factored into some blowing ops followed by some blowing downs which is the riski's result about how bi-rational morphisms can be factored into blow ops and blow downs okay that's all I want to say for the moment about minimal surfaces so the next lecture I will probably be talking about rational surfaces which are the simplest class of surfaces and historically the first ones to be investigated


these students, I'm going to present to you the concept of the joint cumulative distribution function. This concept arises when we are not dealing with a closed variable, but two, three, four or more. So we are actually dealing with a multivariate or a multi-variate situation. Given the random variables X1, X2 and so on that are defined on a probability space, the joint probability distribution for these variables is a probability distribution that gives us the probability of each of these variables falling in any particular range. May the variables be discrete or may they be continuous. Let us concentrate on that case when we have only two variables X1 and X2. In this particular case, we will say that we are dealing with a bivariate distribution because obviously, that stands for two and the joint distribution that can be expressed either in terms of a joint cumulative distribution function, which can also be called a joint CDF or it can be expressed in terms of a joint probability density function. Of course, if the variables are continuous or a joint probability mass function, if the variables are discrete, so students, let us focus on the joint cumulative distribution function and what is the formal definition if we have two variables. The definition is as follows. The capital F of the random variables capital X1 and capital X2 defined at the values small X1 and small X2, this capital F is equal to the probability that the random variable capital X1 assumes a value less than or equal to small X1 and the random variable capital X2 assumes a value less than or equal to small X2. For all values, all ordered pairs small X1, small X2 that are defined on the two dimensional Euclidean space are square. Actually, whatever I said, maybe it was very complicated for you, although it is very simple. When we say and, then when we express it mathematically, then you know that we use the intersection sign there. So that is why what you see on the screen, that carries the intersection sign whereas when I said it, I said and. The second thing is that at the time you are in a univariate situation and you are simply capital F of X, what happens is that the probability of the random variable capital X being less than or equal to the value small X, but also you say this sentence that for all X belonging to R, obviously when it is one, then we talk about R i.e. the real line, which we can also say X axis, but if there are two, then obviously ordered pair will be X1, X2, so it will not be just the line, it will be the two dimensional plane, which can also be called Euclidean space. So it's quite simple. It's just a natural extension of the concept of the cumulative distribution function in the case of one single variable. Now after this, you note that what we are saying is that capital X1 is less than or equal to small X1 and capital X2 is less than or equal to small X2. So we can say that this is a joint event, i.e. two events are different, so we only say that if we are going to deal with them separately, but if we are talking to them together, then we are talking about a joint event and after this, you have a very important result, which has been mathematically derived. I want to share about the cumulative distribution function and I'm talking about that case only when we have two variables, i.e. the simplest possible multivariate situation, when we do not have three, four, five, but only two variables. So what is this result, my dear students, that I would like to share with you? It is as follows. The probability that A1 is less than X1 and X1 is less than or equal to B1, where A1 and B1 are two real numbers and A2 being less than X2 and X2 being less than or equal to B2, where A2 and B2 are two other real numbers. This probability is given by capital F of X1, X2 at the point B1, B2 minus capital F of X1, X2 at the point A1, B2 minus capital F of X1, X2 at the point B1, A2 plus capital F of X1, X2 at the point A1, A2. This is basically the probability of your X1 lying in the range A1 to B1 and your X2 lying in the range A2 to B2. And I am saying that you can denote this by the intersection, but also students that can also be denoted by a comma. You have often seen that the comma is put in this context, which we are discussing It means and. Now these numbers A1 to B1, the limits given for the random variable X1 and the numbers A2, B2, which are for others, it is obvious that A1 is smaller than B1 and A2 is smaller than B2. So now how will we remember the formula given in the form of capital F? We will try to remember that the first one that I said, F of X1, X2 at the point B1, B2, that upper limit of B1 and B2, that's the one. Then when we are minusing it twice, once A1, B2 and the second time B1, A2. And then when we are plusing it in the end, now it is A1 and A2. That is, the limits defined for both the variables are for the lower limit of A1, X1 and lower limit for X2. I am just giving you a tip as to how to remember this particular formula. So this entire discussion that I have just done with you, this is the concept of the cumulative distribution punch.