 We have been looking at problem of solving ordinary differential equations subject to boundary conditions using method of orthogonal collocations. So this is based on using interpolating polynomial and this interpolating polynomial is used over the domain of interest and this interpolating polynomial is then used to discretize the problem, discretize the boundary value problem. So let us look at quick recap of what we have achieved till now. I have been looking at this problem of a general second order boundary value problem which is d z square psi is a general function d u by d z u and z is equal to 0. Now this holds this equation is supposed to hold over domain 0 z less than 1. So here u is some dependent variable it could be temperature, it could be concentration whatever is the variable of interest. So we have this generic second order ordinary differential equation and then I have to boundary conditions I have to bound this. So these are f 1, so this is my boundary condition 1 and then my second boundary condition is d u by d z. So I have this generic problem of solving second order ordinary differential equation subject to these two boundary conditions 1 at z equal to 0 the other one at z equal to 1. So if I draw this on this domain we have this domain here and we have already set the convention of naming. So we have this domain here this is z equal to 0 to z equal to 1 and then we have a solution we have a polynomial solution interpolating polynomial solution which we have shown. So this is u z is equal to alpha 0 plus alpha 1 z plus alpha 2 z square plus alpha n z to the power n this is the interpolating polynomial which is the proposed approximate solution and we have this convention of deciding or calling solution at certain points which are called as collocation points or the grid points. So in this context we are going to call them as collocation points. So we have this collocation points which are numbered you know z 1 z 2 z 3 in general this is z i this is z i plus 1 this is z i minus 1 and the final one is the final point is called as z n plus 1. So we have this n plus 1 collocation points numbered from z 1 z 2 z 3 up to z n plus 1. So these are collocation points. Now these collocation points need not be equispaced. We have looked at finite difference method in finite difference method we looked at two options one was the grid points as they were called in the finite difference method the grid points could be equispaced they could be non equispaced. In this case though in principle no one stops you from taking equispaced points we are going to look at this collocation points chosen in a particular way these collocation points are going to be chosen at the roots of shifted degenerate polynomials. So these are going to be these collocation points are going to be chosen at the roots of shifted degenerate polynomials are given in the lecture notes. So I just list them here so for example if I take the first order polynomial then so first order polynomial then the root is at 0.5 okay if I take the second order polynomial then the root is at 0.21132 and 0.78868 if I take third order polynomial then I have three roots 0.1127.5 and 0.8873 and so on. So if you look at the standard textbooks you will get these roots of the shifted degenerate polynomials and I am going to place this collocation points okay at the roots of this shifted degenerate polynomial which means if I happen to choose three collocation points in this domain okay the first one of course Z1 will be 0 the second one will be placed at 0.1127 okay this is scaled this domain is scaled between 0 to 1 typically if it is length you can divide by length and scale it to 0 to 1. So at 0.1127 will be my second point my third point will be at 0.5 my fourth point will be at you know 0.8873 and by the last point the fifth point will be boundary Z equal to 1 okay so likewise in here in the accompanying notes I have listed roots up to seventh order okay and you will get if you want to know about higher order polynomials you will get that in literature so this but typically it suffices to use third fourth fifth or sixth order polynomials and if you want to have more collocation points then typically what we do is we do orthogonal collocation and finite elements which means we divide it into sub elements and on within that element we define collocation points okay so we don't really go for if you want 50 collocation points we don't do it by taking the 50th order polynomial we take a fifth order polynomial divide this domain into 10 segments and place roots inside each step domain okay but that we will not be discussing now we will be looking more at you know a single polynomial being chosen right okay so let's do a quick recap of what we have done till now where we want to find the solution approximate solution and then this approximate solution I have a naming scheme so my u1 corresponds to u that is this approximate solution computed at z1 okay and u2 is you computed at z2 at the second collocation point okay and so on so in general ui corresponds to u at z equal to zi okay now what we said in the last lecture is that you know we would like to express we would like to express these coefficients alpha 0 alpha 1 alpha 2 alpha n okay these are unknowns okay this is a proposed approximate solution for this ordinary differential equation okay and this coefficients this coefficients I want to express in terms of u1 u2 u3 u4 and so on okay so what is known to me here what is known to me here is I have chosen the collocation point so the collocation points are known to me okay so these locations that is z0 or z1 equal to 0 to you know z2 let's say if you write three points z2 equal to 0.1127 these locations are known to me okay now what we have done in the last lecture is like this okay to get unknowns transformed from alpha 0 to alpha n okay we wrote this equation at n plus 1 collocation points so I wrote this equation u1 u2 up to un plus 1 so un plus 1 is at the last point okay then this will be 1 z1 z1 square up to z1 raise to n 1 z2 z2 square up to z2 raise to n and so on okay and we have this 1 zn plus 1 alpha 0 alpha 1 up to alpha n okay I wrote it in this form this is my I define this matrix as a matrix if you recall I call this as vector theta and these are my unknowns these were represented as capital U okay so to eliminate to express this alpha 1 alpha 2 alpha 3 alpha 4 in terms of unknowns u1 u2 u3 u4 well that the problem here is that u1 u2 u3 u4 are actually solutions of this ordinary differential equation okay at the collocation points so this u1 u2 u3 u4 are not known to us they will be known to us when we solve the differential equation okay so here there is a trouble trouble we do not know u1 to un plus 1 we do not know alpha 0 to alpha n but we want to transform the problems from these as unknowns to these as unknowns okay so the way it has been done is to write this equation u is equal to a theta okay this implies that theta is equal to a inverse u okay theta is equal to a inverse u next what we have done is we need we need derivatives of this function to be evaluated see because here in this equation you have d2 u by dc square you have dU by dc so I need these two equations to be evaluated right so for this what I have done is so in the last class I derived this expression just to recall I will not derive it again I just write the expression okay so we said that we wrote this we wrote as 1 z z the power n into theta we wrote this as in the product of two vectors one vector is 1 to 1 to zn okay and times theta what is theta vector this is theta vector okay this is theta vector and then then we said that this is nothing but 1 z z to the power n into a inverse u theta was replaced by a inverse u okay so in a instead of unknown as theta we have no unknown as u okay and then I wanted dU by dc okay I wanted dU by dc if you take dU by dc this vector becomes z becomes 0 1 2 n z to the power n minus 1 a inverse u right this vector is just 0 2 yeah and then I want to evaluate the derivative I want to evaluate the derivative at these collocation points so if I want to evaluate the derivative at the collocation points this becomes so dU by dc at z equal to zi okay this is equal to 0 1 n zi raise to n minus 1 a inverse u okay so this is the expression that we derived for the first derivative okay similarly we derived the expression for the second derivative okay we derived the expression for the second derivative what is this expression so so I called this okay before I move to the second derivative this this vector okay I called this vector as si I called this vector as si transpose okay this si transpose okay was defined as 0 1 this matrix this a matrix is known to us okay this a matrix is known to us so a inverse is known to us a a matrix can be computed using you know since we know z1 z2 z3 we can compute a matrix we can compute a inverse okay so this this matrix is known since we know zi we also know this row okay so this row times this matrix this will be a row vector that row vector I am calling it as okay si si transpose so my d okay my equation assumes the form dU by dc equal to si transpose u okay so likewise I also derived expression for the second derivative d2 u by dci okay dz square okay and that turned out to be I just give the final expression because we have done it last time so it turned out to be 002 okay we derived a generic expression for the second derivative okay just just just look here this is this is the first derivative if I differentiate this with respect to z again you will get 00 and minus 1 will come here okay this is just differentiation of this this is a constant matrix so so thus this expression naturally follows from this and this is what this is what we got we decided to call this particular vector as ti and decided to call this particular vector as ti transpose into u okay so which means I have an expression that is d2u at zi dz square is equal to ti transpose into u okay I have expression so so what have I achieved what I have done is I have expressed the derivative at a particular point at z equal to zi okay as a vector times unknowns what are the unknowns unknowns are unknowns are values the dependent variable takes at the collocation points u1 u2 u3 u4 okay so this is the derivative at this point is some linear combination is some linear combination of this vector u12 okay see what is the difference here so whether u1 to u2 and u3 define so this is u1 this is u2 this is u3 and so on this is ui ui plus 1 so this will be u so this is zn so this is un this will be un minus 1 and so on okay so what is my u vector my u vector consist of this dependent variable values at point 1 2 3 4 5 6 7 8 okay so this is this is entire vector is that u vector okay see what has happened when you do finite difference when you do finite difference okay you express the local derivative using only neighboring points or you express the second derivative only using neighboring points whereas here whereas here the first derivative or the second derivative is a linear combination of entire okay u1 to un so this is the difference this is the main difference okay everything comes into every comes into the picture okay so this this is much much better way of finding the derivative than taking the local derivative okay and then this now we are going to use to formulate the problem okay so last time we stopped here let's now actually form the you know the substitute these values at the grid points and come up with the equations that need to be solved to get this u1 to un plus 1 we still do not know what are the values of u1 to un plus 1 all that we have achieved in now is to express these derivatives in terms of the unknown variables what are the unknown variables u1 u2 u3 up to un plus 1 which are values the dependent variable takes at the collocation points okay now let's see how to solve the problem now I want to solve this problem okay so actually see this ordinary differential equation where is it defined it's defined on the domain yeah it's defined on the domain 0 to 1 so it should hold where should it hold everywhere ideally the true solution okay should set the true solution using okay this is not a true solution this is an approximate solution the true solution should actually hold at every point in this domain that's what it says okay now when we solve it by orthogonal collocation we are going to say that well the approximate solution should hold only at the collocation points okay right now we are not saying anything what happens in between at the collocation point this equation should be satisfied at the collocation point this equation should be satisfied okay so this is discretization actually actually we are looking at only finite number of points where the equation should hold the original equation should hold everywhere okay so what's going to happen now is because of this approximation this ordinary differential equation will get transformed into set of non-linear or linear algebra equations okay then so this should hold inside the domain at the boundary point what should happen this equation should hold at first boundary this equation should hold at the second boundary so this these equations okay enforcing this equal to zero at each of the collocation points will give you a set of equations plus these two will give you two more equations will have number of equations equal to number of unknowns until we are going to solve it okay so now let's let's embark upon solving the problem so the game plan is is enforce so this is called residual I will just write down what you will understand what why I am solving the residual to zero at each collocation point so that means what I am going to do I am going to solve this equation this is called residual residual psi okay now d2 u zi by dz square du zi by dz u zi okay or in in our case u zi is nothing but u i zi should be equal to zero this this term I am going to call as residual I want this equation to hold at every grid point which are those grid points for i equal to 2 3 up to n okay I want this to hold at okay now if I substitute if I substitute for expressions that I have got see that the derivatives are now expressed in terms of algebraic expressions okay I am going to replace them so this actually means I want to solve for ti transpose u si transpose u u i zi equal to zero look at this equation here okay see du by dz square I have replaced by equivalent algebraic approximation okay sorry d2 u by dz square I have replaced by appropriate algebraic approximation du by dz I have replaced by appropriate algebraic approximation okay so this differential equation is now converted into an algebraic equation how many such equations we have got now we got 2 i equal to 2 3 4 up to n so how many equations n minus 1 we got n minus 1 starting from 2 to n we got n minus 1 equations okay we are going to set this residual equal to zero at each of the collocation points internal collocation points okay now what about boundary conditions see how many unknowns are there u1 u2 u3 up to un plus 1 okay so how many equations you need you need n plus 1 equations to solve get okay how many equations we got till now n minus 1 so we need 2 more equations those 2 equations are going to come from boundary conditions okay so the boundary conditions will give me additional 2 equations that completes the set you get n plus 1 equations in n plus 1 unknowns and then we are solving the problem we will use it okay so let us write the 2 additional equations so the next equation is f1 okay s1 transpose u u1 at z equal to 0 equal to 0 this is my first equation this is my first equation and f2 sn plus 1 transpose u un plus 1 and 1 equal to 0 okay see now now you have these 2 equations these 2 equations okay and this these n minus 1 equations okay these n minus 1 equations together with these 2 equations forms the set of n plus 1 equations n plus 1 equations in n plus 1 unknowns we need to solve them simultaneously okay if these happen to be linear algebraic equations you can solve them analytically if they happen to be nonlinear algebraic equations you have to solve them iteratively using some alternative method okay so what we have done here we have achieved transformation of the problem of a boundary value problem okay from a differential equation to a set of algebraic equations okay using approximation theory what what method in approximation theory we have used we have used interpolation interpolating polynomial okay so this these equations then then you know you can solve it using the standard tools like Newton's method Newton-Raphson or successive substitutions whatever whatever is suitable that you can use to solve this particular problem afterwards okay let's take a specific example then it will be easier for you to understand before that I just for the sake of convenience I want to define 2 matrices okay using this S vectors and using this T vectors using this T vectors I am going to define 2 matrices S and T okay I am going to give you a method to compute them very easily okay so we will define these 2 matrices and then for a given number of collocation points one has to first construct these matrices and then use them to formulate your equation one thing which which I would like to bring to your notice here is that in every equations these are dense equations these are dense equations in every equation okay u1 to un plus 1 will appear okay because the derivatives are approximating not locally but using all the points in the domain okay since the derivatives are you know approximating using all the points these equations will be dense okay so this is this is this is something different from if you do finite difference okay only 2 neighboring variables will appear in one particular equation here it is not like that every variable will appear in every equation particularly if you take a single polynomial over the entire domain okay so that is that is a bit difference okay so let me define this matrix S and T and then we will take this particular example that we have been using quite often or we will be using quite often in the course which is a similar reactor with axial mixing so before we do that let me define these matrices okay so this S matrix is going to be defined like this it is it is it is consist of S1 transpose S2 transpose so here S superscript 1 implies it is a vector first vector second vector third vector and so on so this is this is going to be S n plus 1 transpose okay so this is a n plus 1 cross n plus 1 vector okay this is a n plus 1 cross n plus 1 vector okay and it is very easy to show that this can be computed by this can be computed by looking at this another matrix 0 1 so if I if I decide to call this if I decide to call this matrix as say C matrix then this is equal to C times A inverse C times A inverse well how was our A matrix defined A matrix was defined just keep it in background that we have this equation that U is equal to A theta okay U equal to A theta so A matrix is defined okay using 1 Z1 Z2 X1 so likewise likewise I also want to define this T matrix I am going to define this T matrix this T matrix will consist of okay so I have stacked up I have stacked up these row vectors I have stacked up these row vectors okay to create this matrix okay why why this for the notation because in our course whenever we are defining a vector it is a column vector when I want to make it row vector I am taking a column vector and putting it as a transpose that is why this notation which is which we are getting okay and then you can show that this matrix is equal to this can be very easily computed using D times A inverse and what is this D matrix okay this is also n plus 1 yes this is also n plus 1 cross this is also n plus 1 cross n plus 1 matrix yeah so let's write this D matrix this D matrix is slight modification of this matrix so D matrix will look like this D matrix will be 0 yeah 0 0 this will be 0 0 2 6 Z1 this is n minus 2 this is n minus 2 and so on so this is nothing but this D matrix will of this stacked vectors okay these are nothing but TI vectors evaluated at different collocation points and this together with A inverse post multiplied by A inverse is going to give me T matrix okay so I need to create there is a preparation for solving this problem okay I need to create this S and T matrices okay by choosing collocation points once I choose collocation points okay 3 4 5 6 whatever once I choose the collocation points I can first find out A matrix once I find out A matrix I can find out A inverse and then I can define C and D matrices and from that I can get S and T matrices once I get S and T matrices I want to use rows of these matrices to discretize my ordinary differential equation and convert it into converted into algebraic equations let's take a specific problem then you will understand it better okay so this is our tubular reactor this is tubular reactor with axial mixing okay this is the problem which we have in which we have looked at earlier when we studied the finite difference method okay the associated ordinary differential equation is 1 by Peclet number into d2c by dc square minus dc by dz minus dac square is equal to 0 so this is the ordinary differential equation that should hold between 0 z1 then you have 2 conditions okay you have 2 conditions that is dc by dz okay is equal to Pe into c0 minus 1 this should happen at z equal to 0 and dc by dz equal to 0 equal to 0 okay this is the second condition that we have 2 boundary conditions and second boundary condition should hold at z equal to 1 the second boundary condition should hold at z equal to 1 so this is my problem this I want to discretize okay let's say let's say I have chosen 3 internal points okay so I am going to do a very simplistic solution so this is my z1 this is my z5 okay I have 3 collocation points okay this is z2 z3 z4 okay I have taken collocation points at the roots of the third order degenerate shifted degenerate polynomial okay so this happened to be you know so the first one is at 0.1127 second one is at 0.5 third one is at 0.8873 okay this z5 is equal to 1 and z1 is equal to 0 so we have 5 collocation points 3 internal collocation points 2 boundary points okay and then we are going to get 5 equations in 5 unknowns okay we are going to get 5 equations in 5 unknowns now if I okay what is the first thing to do here first thing to do is to compute a matrix okay first thing to do is to compute a matrix so a matrix okay if I actually compute a matrix okay if I compute using these 5 points it would be 5 x 5 matrix okay then if I compute S and T matrices okay I will just write here a sample of S and T matrix so S matrix first row turns out to be minus 13 14.79 minus 2.67 1.88 minus 1 and so on okay so this is a 5 x 5 matrix so there are 5 rows this is the last row this is the first row okay likewise knowing these points once I have chosen these points I can compute a matrix okay then I can compute C and D matrices then I can compute S is equal to C into A inverse I can compute T is equal to D into A inverse because AC and D these matrices depend only on the values of the collocation points okay so once I have got these matrices okay then I am going to use these rows of these 2 matrices to convert this differential equation into set of algebraic equations okay so what is my first equation so I have 3 equations at the 3 internal collocation points coming from this differential equation I have 2 equations coming from boundary equations okay so my equations would look something like this okay so 1 by Pe into Ti transpose times C vector minus Si transpose times C vector minus DA Ci square is equal to 0 here okay I going from 234 okay what is the C vector? C vector now consists of C1, C2, C3, C4 and C5 okay C vector consists of C1, C2, C3, C4, C5 okay so this row times C1, C2, C3, C4, C5 okay plus this row times C1, C2, C3, C4, C5 plus C2 square when I equal to 2 okay second row here you will choose the second row from the matrices here when you choose I equal to 3 you will choose the third row from S and T matrices okay and you will get here DA into C3 square okay so you are getting 3 non-linear equations you are getting 3 non-linear equations and the 2 additional equations arise from boundary equations okay so the 2 additional conditions that you get is S1 transpose C minus Pe into C1 minus 1 is equal to 0 and S5 transpose C minus is equal to 0 okay so these 3 equations these 3 equations plus these 2 equations together they form 5 non-linear algebraic equations which need to be solved simultaneously okay because particularly these equations everything appears all the equations C1 to C5 will appear in all the equations okay these are coupled non-linear equations they have to be solved iteratively using Newton's method Newton's Raphson or some non-linear equations all the which may be optimization whatever is at hand for you to solve this so this is the original problem okay which is actually defined on a domain which is not finite dimensional see the true solution here let's come back here what is the true solution here true solution here is a concentration profile as a function of Z okay Z varying from 0 to 1 so it's a function okay it belongs to which set does it belong to it belongs to the set of continuous functions twice differentiable defined on domain 0 to 1 the true solution is actually that we have discretized the problem okay using interpolation polynomial only converted into a fifth order you know fifth not sorry 5 dimensional vector okay so we are approximating an infinite dimensional solution using the 5th 5th order sorry 5 dimensional vector well you can increase you can increase the number of colocation points to 7 8 9 but you know how many you can go so if you want to really make a higher dimensional approximation what one could do is one could divide this into segments and on each segment one can define a lower order colocation polynomial okay then of course that is called as ortho collocation on finite elements then you have to write conditions by which the neighboring solutions are you know meet each other so those those conditions will have to be written okay additional conditions will come to maintain the continuity of the solution you will need additional conditions to be imposed okay but this is the basic principle once you understand this you know extending it to finite element is not difficult this concept is this okay now before we close this lecture I also want to show that this is not just converting partial the boundary value problem I am going to just take up a version of the same problem which is a partial differential equation okay and then you will see that the partial differential equation okay will get converted into a ordinary differential equation set of ordinary differential equation here here this is a there is no time involved here okay I am going to now convert this into a partial differential equation by including the time derivative if I put a time derivative the same you know same problem okay instead of getting transformed into set of coupled algebraic equations non-linear algebraic equations it will get transformed into set of coupled ordinary differential equations okay then then of course you have to use methods to solve the ordinary differential equations that is a separate thing okay right now we are just looking at a problem transformations okay so let's do a quick quick recap what we have done is we have this we have this second order differential equation okay we wanted to we have proposed a polynomial interpolation based solution approximate solution for this dependent variable in terms of independent variables e okay and then we are forcing this residual we call this as a residual to be 0 at a finite number of collocation points these collocation points are chosen at at roots of the shifted Legendre polynomial okay why why shifted Legendre polynomial why not why not at some you know say regular intervals and so on it has been found that if you actually place them under shifted Legendre polynomial then the approximation errors are no okay so so the reason for choosing or though normal or orthogonal polynomial roots of the polynomial is to get less approximation errors okay so so there's a reason why we choose the the collocation points in a special way and not increase space and so on so let's not get into that part but just accept this now that you put them at special locations then the approximation errors are no okay so then we looked at one problem which is together reactor with axiomixin and this problem what has happened is we are able to convert this particular problem into set of five coupled nonlinear algebraic equations which need to be solved iteratively further okay which will give you an approximate solution okay now here because the derivative approximation is much better okay typically it is found that a good solution can be obtained using less number of collocation points so till now we have looked at finite difference method okay in a finite difference method you need large number of collocation parts sorry you need large number of grid points to get a good solution because you know you are taking local approximation of the derivative okay so you need large number of grid points to get a good solution so what is the meaning of large number of grid points large number of grid points means see suppose you were to enforce this equation at this residual to be equal to 0 at large number of grid points the number of equations that you need to solve simultaneously will be large suppose you know to get a good solution using finite difference I need to subdivide this into 100 small intervals so then here you know when you transform this into algebraic equations you get hundred algebraic equations plus two algebraic equations at the boundary so hundred two algebraic equations okay what is found here is that with this approximation less number of collocation points a smaller order polynomial gives you a good solution in many cases okay so here using this method you can get good approximations using you know less number of less computations in some sense okay so just to compare let me before before we just close the lecture let's just look at the partial differential equation and then let's see what happens now I'm going to keep the same problem I'm not going to change the problem except in this case in this case I looked at the steady state solution I did not involve I did not consider time but suppose you were to consider the transient response of the tubular reactor with axial mixing let's keep the same problem okay so I'm going to say here dou c by dou t is equal to so these become partial derivatives dou 2 c by dou c square dou c by dou c into so the time derivative is earlier we had put it equal to 0 now the time derivative is not equal to 0 so now my second example that is this is pd e okay so I want these conditions I want these conditions to hold at all time so I want this condition to hold at all the times so ct by dz okay is equal to okay now the time has come into picture okay and I want the solution to obey these I want the solution to obey these equations at all the times okay at all the times so this is ct so now there are two attributes to the solution time and space that we are looking at a boundary value problem the only one attribute that is one independent variable that was space now my solution will be time and space okay so when I convert this problem when I discretize this problem I'm going to only consider discretization in space I'm not going to write out discretize in time I'm going to keep time intact okay and discretize only the right hand side the spatial part okay so what happens if I so so at the internal collocation points at the internal collocation points I get three differential equations what are those three differential equations now there are three differential equations in time okay the right hand side becomes algebraic okay because these derivatives using method of collocations okay this we don't approximate okay right now so we have this equation which is written at the three internal grid points okay dci by dt okay is equal to 1 by x number into ti transpose c so here now of course ct minus si transpose ct minus dacit square okay I going from 3 and 4 okay I going from 2 3 4 okay and then this last two equations actually last two equations becomes two algebraic constraints so what you get what you get is set of differential algebraic system okay what you get now because these these these are spatial derivatives these are this is a time derivative okay the special derivatives will get covered into algebraic equations okay so there are algebraic constraints to be obeyed at the boundary points okay and this partial differential equation gets converted into ordinary differential equation okay so we have three ordinary differential equations and there are two algebraic constraints at the boundary that is s1 transpose ct equal to okay so there are three differential equations these are three coupled nonlinear differential equations and they are coupled tightly with these two algebraic constraints and they have to be solved simultaneously you can see here what is the ct vector ct vector is c1t c2t c3t c4t and c5t okay so dc2 by dt dc3 by dt dc4 by dt all of them are functions of c1 c2 c3 c4 c5 not only that because of this square coming here these are nonlinear functions okay so there are three nonlinear differential equations two algebraic equations coupled okay in this case you may be able to eliminate two variables let's say you decide to eliminate the the first and the last okay you decide to eliminate you you decide to eliminate c1 and c5 it will be possible because these two are linear constraints these two are linear constraints okay there is no nonlinearity here so you might be able to rearrange and express okay c2 you might be able to express c1 and c5 in terms of c2 c3 c4 if you do that then you can eliminate here and then you will get three differential equations in three unknowns and then you can solve them by whatever method but then but then you know that is one way or you you solve them simultaneously using a method for solving differential algebraic systems okay so this is how you transform a problem which is originally ordinary differential equation boundary value problem okay into set of algebraic equations linear or nonlinear okay if it happens to be a partial differential equation you will get a set of ordinary differential equations plus algebraic conditions these have to be solved simultaneously and then you arrive at the solution okay so what we have learned in this in this part is that you know how interpolation polynomials can be used to transform a problem okay so we began by see what is the foundation of all this the foundation is that you know why why we could construct a polynomial solution because sometime back I talked about based on theorem what does based on theorem tell you yes last year tells you that any any continuous function can be approximated arbitrarily by using a suitable order polynomial okay that's why we could we could construct a polynomial approximation to the solution c z or c t z okay that's why we could construct a polynomial approximation okay now using the polynomial approximation how do you construct a polynomial approximation based on theorem only tells you that there exists a polynomial approximation here you know you actually can you have to actually construct it the first method that we saw was never see the approximation okay that led to finite difference method the second method that we saw was interpolation that has given rise to orthogonal locations okay in the next class onwards we will start looking at these squares method so these square fitting okay and then in the in the context of these square fitting is a very very vast area and I will be talking not just about converting boundary value problems of partial differential equations I will be talking about many more things under these squares now there you know in the context of partial differential equations of boundary value problems will get the method of finite element this is called method of finite element okay so with this we will close this lecture and move on to these squares methods in the next class