 So till now we were looking at concept of vector spaces or fundamentals of vector spaces and we revised many concepts that are very useful in subsequent development where we will be talking about solving problems numerically. So we started with concept of a general vector space we qualified sets which can be denoted as vector spaces in the sense that of the three dimensions of the three dimensions then we went ahead and talked about concept of let not which is generalization of concept of length. After that we looked at enough vector space in which motion of angle business and we had orthogonal space. So the nice things that we like in three dimensions have now been made available in any other vector space. You have orthogonality, you have orthogonal functions, you have orthogonal sets, you have orthogonal polynomials and now we are poised to start using them to solve problems numerically. Well what kind of problems in my first lecture I talked about the classification of problems and also on the Moodle webpage I have uploaded one module, module 0 which talks about different problems in chemical engineering now most all of you are post graduate students so I have not really talked about those problems on the board, we will be meeting problems as we go along through assignments or even through when we do some developments. So physical problems you already know where you get partial differential equations, where you get set of algebraic equations, non-linear algebraic equations, ordinary differential equations and so on. So these things I am assuming that you already know to some extent that is the motivation from the physics or chemical engineering point of view where do these problems arise. My motivation or my now next aim is to look at problem transformations. So first of all what I am going to do is using the language of vector spaces that we developed in now I am going to represent different equations that you are familiar with or that we are going to solve as a part of this course and then I am going to show you that actually we cannot solve them in most of the cases exactly analytically we have to construct numerical solutions but is numerical solutions are not in the same space as that of the original problem. So the spaces associated with the original problem and the spaces associated with the solutions are different so intrinsically we are finding an approximation to the truth and you should be aware of this reality. So my second module which will almost be about 12 to 14 lectures will be devoted to problem discretization. Now approximation theory is a branch of applied mathematics and we obviously cannot do a justice to this in few lectures we are just going to see only some part of it not everything whatever is relevant to this course but between the problem which you are defined from physics your real world problem and the solution stands this approximation theory we cannot in most of the cases we cannot solve the problems analytically we have to resort to some kind of numerical approximation and so what is involved what is conceptually happening when you go from the original problem to the transform problem. First of all can we represent all these problems now that we have defined this new concept of space can I generalize my idea of you know how a problem is in general defined in applied mathematics can I say that all these problems are in some sense the same problems can I classify these problems what kind of problems that can arise in applied mathematics or in numerical analysis for solving integral problems. So let us look at this classification and then so we will get an overview bird's eye view and then we will start looking at specific solutions. So everything that we develop till now you will start seeing its applications as we go along in next 10 to 14 lectures okay. So first of all you should know that you have an original problem here and then you use approximation theory and then you get a transform problem and this transform problem is something that we solve using different numerical techniques okay. So we apply different numerical techniques or numerical tools I would say so you use different numerical tools and then you solve an approximate numerical solution. So this is very very important this is very very important and if you get a bird's eye view of what is really happening in approximation theory then given a problem original problem you will be able to think of a transformation there is no unique way what you will realize when you do this transformations but I will take the same problem and show you how it can be transformed in multiple ways okay. I will start with a boundary value problem I will show you that there is you know transformation A transformation B finite difference method orthogonal collocation method there is different ways of transforming the problem and you know finite element method and so on. So same problem can be approximated in different ways when each one of them has advantages and disadvantages there is no one way which is unique in such that you know it takes case of all the difficulties there is something to be given and something to be taken so there is always a give and take. So you get a transform problem and then you use the known numerical tools like algebraic equation solving non-linear algebraic equation solving and get a approximate solution. Then well as a chemical engineer or the physicist you should go back and see whether the physical solution that makes sense here so many issues when you do from here to here for example the transform problem could be ill conditioned the solution that you get may not be a good solution because of you know ill conditioning problems and so on. So just because I have a very good computer and I have a good you know powerful program with me to solve it does not mean that my solution which I get is finally correct there could be problems on the way could be problems here there could be problems here okay. So again for some time you will see generic things and then we will start getting into specifics my first task is to define a transformation so what is the transformation transformation is any operation or any rule that takes an element from a space X and transform it into an element in space Y okay. So I have here two vector spaces I have a vector space X or vector spaces X and Y are vector spaces M is some subset of X well I am still writing generic things but as we go along all these ideas will become fall in place will become clear what it means in concrete terms in terms of so transformation is a rule that associates with every X that belongs to M and element Y and element Y which is okay I have not written the words but the bare definition if I am given two vector spaces X and Y and I take a subset M belonging to X okay and then if I pick any element X then a transformation is a rule that gives me element Y from space Y okay gives me an element from space Y before I move on to give you examples of such transformations well you are aware of some transformations and then I will show you that everything that we need to solve as a part of this course actually can be put into this generic framework okay before I move on I am going to define two concepts one is a linear transformation okay one concept is a linear transformation so linear transformation is a special class of transformations not all transformations are linear okay linear transformations if I give you X and Y that belong to set M we are still talking about the same set M subset of X and so on and we have defined this transformation okay and if I pick up any two scalars alpha beta say belong to set of real numbers or the field on which you are working with let us for the time being take real numbers okay so if I can write transformation of alpha X plus beta Y is equal to alpha times TX plus beta times TY if I can do this if I can do this if a transformation allows you to write transformation of this combined vector alpha X plus beta Y as alpha times transformation of X plus beta times transformation of Y okay then such a transformation would be a linear transformation give you a simplest example let X be RN and Y be RM okay Y is equal to AX where A is a M cross N matrix M cross N matrix is this a linear transformation just apply the definition I take A times alpha X so let me take two vectors say V1 V and U that belong to X I am taking two vectors V and U that belong to X okay what will happen to alpha U plus beta V times A I am constructing a new vector alpha U plus beta V alpha and beta two scalars okay alpha beta R2 scalars now this is equal to alpha times AX plus beta times sorry AU plus beta times A right AV is this a linear transformation just apply these definitions T see this is a transformation Y is obtained when A operates on so what is the operator here T transformation is A operates on X give any element X A will operate on it and give you a element in it will take an element from N dimensional space will give you an M dimensional vector okay so this is a linear transformation it is a linear transformation right now tell me whether this is a linear transformation the next one well my X is still RN my Y is RM okay I am defining a transformation Y is equal to AX plus B where B is a vector is a constant vector just do it is it a linear transformation who said yesterday why plus B so if you actually take a vector if I take alpha V plus beta U I would write I would write when I am operating on this vector I would write like this isn't it Y is equal to okay now this part cannot be split as sum of two transformations I cannot write this I cannot write this as what will be what will be what will be T U T U Y is equal to transformation of U or Y is equal to transformation of U will be will be AU plus B okay what will be transformation of V AV plus B okay AV plus B so for this particular transformation I cannot I cannot split I cannot write T of alpha X plus beta Y is equal to well well one mistake which I have made here in notation let me correct it let me correct it here well let's not take this X and Y let's test this U and B otherwise you will get confused between this Y and Y here okay so U and V here and let's write U and V here so U and V are two vectors from M not X and Y because we are using Y to denote the element in the range space what is range space which is the domain space X is the domain space M is the domain and Y is the range space so now let's U and V are the elements from M okay and Y we are using to denote the range space so just this definition stands corrected so now so what actually would appear to be a linear transformation is not really a linear transformation you cannot you cannot satisfy the basic condition of linearity for this particular transformation okay now everything that is not a linear transformation any transformation that is not a linear transformation is a non-linear transformation the simple definition okay whichever transformation does not obey this simple this simple law when you apply operator T on alpha U plus beta V you should be able to write alpha times T operating on U plus beta times T operating on V it is not a linear transformation okay so now let me give you some more examples of the transformations and I just want you to think that understand that these new examples that we are going to talk about are not different from so which is the equation which is probably most familiar to you the most familiar equation that you have from your undergraduate or from right from your 12th standard 11th standard is this or may be now it is introduced in ninth and 10th standard right solving linear algebraic equations so well after sometime you realize that it is nothing but a matrix equation so you are taught about a matrix so what is a matrix matrix is an operator operating on X giving you an element Y in the range space what I want to show you is that after we have generalized the concepts of vector spaces almost every problem that we encounter in engineering can be viewed in the same manner it is not different okay it is an operator operating on a vector giving you another vector in the product space in the range space okay so now let me move to the second example maybe say third example so one example I give you was AX plus B so my third example here is so my X here is C1 set of continuously differentiable functions on interval AB set of once differentiable functions set of once differentiable functions on interval A and B so independent variable T belongs to AB so YT is equal to D by DT of XT okay I am taking a vector from set of continuously differentiable functions once differentiable functions operating this operator D by DT on it this is my operator T it gives me another vector Y which is from the set of continuous functions the resultant need not be differentiable okay I am getting a set of continuous functions okay so this is space this is my space X this is my space Y this is the operator T which operates on a vector this equation is fundamentally not different from Y is equal to AX A is a matrix which operates on a vector X gives me a vector Y no difference here I am getting a continuous function when operated by a differentiation operator on you know a vector what is this vector this is a differentiable function on define on some interval could be whatever it could be 0 to 1 it could be 0 to infinity it could be minus infinity depending upon your application of your choice the interval would be different but conceptually Y is equal to AX and this equation are not different they are one and the same and so is the case if my operator is little bit different if I write this is third another example YT will be 3 this is another operation this is another operator T 3 times D by DT plus 5 so actually you will get 3 DX by DT plus 15X this is an operator operating on this vector giving me another vector okay giving me another vector no difference it is an operator exactly in the same sense as okay now here you are going from n dimension to m dimensions m could be smaller m could be larger depending upon what kind of matrix you are talking about when you deal with equation matrix equations let me give you another example of transformation in fact when you start understanding this language of vector spaces very very powerful language developed in the beginning of 20th century or end of 19th century everything you know falls logically into the place you can see start we will start seeing you know one unified structure in all the mathematics applied mathematics that you have been studying okay now let me look at this next example you know my X so AB is some interval and I am taking set of all integrable functions I am taking set of all integrable functions on the interval and then I am defining a transformation alpha is equal to a to b what is this you are familiar with this definite integral area under the curve so what is the what is the range space R right so this is an operator so I would say that this operator a to b something DT this is my operator T this is my operator T it takes a vector from the space and maps it on to R okay so it is a function from let us call this whatever this X which is defined here okay X to R this operator T operates on this vector okay gives me one number gives me a one scalar number okay so this is from see just like in this case you can have an operator A which is from higher dimensions to lower dimensions okay you can have operator A which is only where Y is a scalar you can have that right it is you can have an operator which will give you as a scalar value R in R okay so likewise this is an operator which takes a function from this infinite dimensional vector space and gives you a scalar value in R it is a transformation is it a linear transformation so what will happen if I do if I do AFT plus BGT DT no not AB we are taking AB here no we will take delta and gamma or the two scalars and this goes from A to B so I can write this as delta times AB FT DT plus gamma times integral over AB GT DT okay so this is delta TFT plus gamma TGT everyone with me on this so this is a linear transformation this is a linear transformation I can write this as a linear combination of two vectors in the product or in the range space okay this solution plus this solution they linearly add up so linear transformation in the same sense this is a linear transformation in the same sense this is a linear transformation try to understand this okay no difference between what you have done here and what you have done here all these are linear transformations so if linear transformations have some nice properties they not only hold for finite dimensional vector spaces you can translate them into any other vector space that is the power of using the vector space approach okay so far so good let us move on to the problems that we encounter more frequently one is boundary value problem can I represent it in the new language of spaces can I view it as a transformation from vector space to another vector space how will I represent this I will do a little bit more work we have to go back to the definition of product spaces but still you can do it you can show it is a transformation from one particular type of one domain space to another range space okay so this particular AX equal to BY is equal to AX I am not going to wipe it out I am going to just leave it this is the reference we know this very well and I just want to map everything to this particular what about ODE IVP ordinary differential equation initial value problem can I represent it as a transformation okay well my ODE IVP is given by say DX by DT is equal to F of XT T where F is some this kind of equations we encounter say solving a batch reactor problem or batch distillation problem so we will have equations which are of this type DX by DT is equal to F of XT where X are states and then T would come because you are giving some input policy like cooling policy or heating policy and so on so this function of T will appear so this kind of equations we very very frequently have to use let us look at right now one dimensional equation so one dimensional means there is only one state let us not complicate life by getting into n differential equations in n unknowns which will be anyway hitting into later but right now for representation purpose let us look at so what should X belong to the solution X where should it belong to C1 she says C1 do you agree what is C1 set of once different stable functions so the solution X should belong to the set of once different stable functions is it clear it has to otherwise you know the equation is not defined okay so my solution X so my Y should be now this independent variable T it belongs to some set AB okay so why should be why should be but this problem it comes up we are normally given we are normally asked to solve this problem for some initial condition at XA right we are normally asked to solve this problem for some specified initial condition XA at point A let us say that is the initial point of the interval if it is time 0 to infinity at time 0 you specify what is the initial condition and then you want to solve this problem from T 0 to T infinity okay so you are typically given so that is why it becomes initial value problems ordinary differential equations initial value problem okay so the solution actually belongs to the product space R just think about it where R appears because of the specified initial condition the solution should satisfy two things it should be once differentiable function and initial value should be equal to R initial value should be equal to XA XA is some number from real line we are talking of real differential equations real valued functions we are not getting into complex okay real valued functions so the first value initial value is a real number okay so the range space is nothing but set of one differentiable functions the product space formed by this and R R comes from here okay and what about X so domain is well I must check out here whether this whether Y also has to be once differentiable yeah I think I think the description is correct so my X is set of one differentiable functions my Y is a set which is once differentiable and cross R so this is a map this is a map from so this transformation so what is this transformation how will I represent this so I would represent this as D by DT minus F operating on XT operating on XT see this XT operates DX by DT F operates on X you will get F of XT okay F is the operator which operates on vector X okay and what should be the right hand side right hand side should be 0 vector and the initial condition if you ask me to represent this as a transformation in the same sense as Y is equal to AX I would write it like this okay this operator this operator what is this operator DX by DT minus F of X F of XT okay given any vector this operator operates on X okay when I take this on the left hand side what is on the right hand side but it is not 0 it is 0 vector is 0 vector okay is 0 vector in this space set of continuously differentiable functions and what should be the solution what is the requirement that the solution should have for the solution that initial condition should be XA okay see this is my vector in the cross space in the cross space here okay so this operator operating on X should give me 0 vector plus the initial value of X should be this is not X do not confuse this 0 to be X this is not X see when I transform okay look let us look at it okay see this is D by DT minus F this whole thing when it operates on XT it gives me DX by DT minus F of XTT this is what I get okay but when I have taken this F on the left hand side okay what is on the right hand side this is equal to 0 but 0 means 0 where 0 this is 0 for over the entire interval this is not one point right see this is a vector XT is a vector defined from A to B okay so if this is satisfied if the solution satisfies this equation I should get 0 everywhere on the interval A to B so this is like solving you know operator operating on the vector gives me 0 we have solved this kind of problems right we have solved AX is equal to 0 we solve this kind of problems AX is equal to 0 right when columns of A are linearly dependent you can get a solution we saw that we saw took an example we got non-zero solutions okay so those of you who have been by now introduced about null space and range space you would know that the solution belongs to the null space and so on so this is a problem which is similar to that operator operating on X gives me 0 vector so that is why here I am writing 0 vector and X is addition should have initial value to be equal to XA so this is a representation of so same thing is happening like this problem okay see we solve for AX is equal to B where B is a specific vector right this is a specific vector here 0 we solve for AX is equal to B okay in fact I am trying to solve for AX is equal to 0 okay the parallel of that is this operator operating on X gives me 0 vector okay yeah because my solution should have initial condition equal to this I want that vector as a solution this is a way of representation I want that vector as a solution for which initial value should be this and the right hand side should be all 0 vector in the equation I take a vector C I want a solution XT such that X okay so let us take instead of let us let us I think this will be easier XA is equal to some alpha given that XA is equal to alpha okay XA is equal to alpha so I will write here alpha I want a solution I want a vector I want a vector XT such that which is solution of this problem such that XA is equal to alpha condition number one okay initial condition should be satisfied by the solution right this is the condition number one what is the second condition if I operate by this operator on this X I should get 0 vector I should get 0 everywhere on interval A and B why because XT is defined on interval A to B okay this is a representation of the problem that let us move on to boundary value problem okay just think about it it may not sink because you know we are suddenly you are being asked to change the gear from three dimensions to some infinite dimensional spaces and operators from any space to any space it may take some time to sink think about this okay think about what I am saying now consider boundary value problem see a typical boundary value problem will consist of a differential equation that holds in the domain say 0 to 1 okay Z going from the independent variable going from 0 to 1 one is dimensionless length let us say and there are two boundary conditions one at the Z equal to 0 other one at Z equal to 1 okay so some boundary condition F2 du by dz at Z is equal to 1 u1 is equal to 0 okay and you make an attempt to write what are the spaces here so this operator operating on u should give you 0 not at one value where it should give you 0 everywhere between 0 to 1 excluding the two boundary points excluding the two boundary points okay now two boundary points you know you can actually specify okay this instead of giving this 0 we will put this as alpha 0 and alpha 1 so I could specify some conditions here boundary conditions that some derivative plus some operator operating on this derivative first derivative and u0 gives me alpha 0 and this gives me alpha 1 so these are these two conditions okay so what is the underlying domain here twice differentiable function on 0 to 1 interval is 0 to 1 okay so my domain space here is C2 0 to 1 and my range space here is my range space here why is this R and R coming to boundary conditions the solution should satisfy this boundary condition at initial point this boundary condition at Z equal to 1 okay and this is the map so what is the map so what is my transformation now my t corresponds to d2 by dc square plus a times plus b one minute here we have not taken any a b so b times d by dc plus c times whatever whatever it operates on so this operates on uz so this is my this is my t operating on uz okay here it gives me 0 vector here I am asking you to find out for solution for 0 vector on the right hand side in general the right hand side here in general right hand side here need not be a 0 vector I I can give you this equal to sin z for example right so then it will be you know this operator t operating on uz giving me giving me sin z whether solution exists or not is a different story but I can ask you to do this okay and now this is same as this is same as y is equal to or ax is equal to b conceptually no difference between ax equal to b and you know this operator operating on this vector giving me another vector which is sin z okay I am trying to solve for ax equal to b okay no difference the same thing so once you have this unified view of vector spaces you can start looking at all these problems are actually the same problem okay so now can I can I say what is can I state a generic problem in applied mathematics that or in another saying applied mathematics applied engineering mathematics well I will I will do a generalization here and say that any problem so likewise here in my notes I have also talked about how to look at a partial differential equation as a transformation from a particular space to on the space domain space to range space you can go through the example here okay so so what is what is the problem that we have to solve as a part of applied engineering mathematics we are given some domain some subset m which is which is which is subset of x and y is range space okay y is range space and then what am I supposed to do I am supposed to find I am supposed to solve this equation y is equal to transformation of x okay such that y belongs to y and x belongs to x x belongs to m which is subset of x so y belongs to the to the range space and x belongs to so this is the problem this is what we have to do and there are three problems that arise from this okay I will which I mean all the problems that we encounter in engineering mathematics can be further classified into three problems direct problem in direct problem operator t is given x is given you want to find out y okay given t and x find y okay given t and x find y example definite integrals okay inverse problem what will be inverse problem just think about y is given t is given you do not know x usual problem a x equal to b b is given a is given x is not given to you okay a differential operator is given right hand side is given what is the vector that gives you the right hand side okay okay and a third problem what can you think of the third problem third classification find the transformation given x yeah example fitting correlations finding out rate equation for in reaction engineering you are given some measurements you know you want to find out the model parameters okay so the third problem is identification problem well will deal little bit more about this in our next lecture so there are three essential problems that are three essential classes of problems that we encounter in engineering mathematics out of which I am not going to be so much worried about the direct problems they are relatively easy to solve what are difficult to solve are inverse problems okay solutions the so called solutions of algebraic equations non-linear algebraic equations boundary value problems partial differential equations all these will fall into the inverse problems we are given a vector in the range space we are given the operator you want to find out a vector in the domain space which satisfies this particular equation okay and a third problem which we are going to look at little more elaborately that is identification problem that is given y and x we want to find out the operator t okay given the observed data input given to a plant and output given to a plant I want to find out the transfer function in process control if you remember you are finding out the transfer function finding out the transfer function is nothing but finding out a differential equation that governs the input and output behavior okay is the problem of identification okay so these two classes I will once over once again briefly go over these classes classification in my next class and then now we will start you know dealing with the last two classes that is inverse problem and identification problem so how is approximation theory going to be used here that is what we start from the next class.