 So, let us consider now a very special case of a second order differential equation right linear differential linear constant coefficient second order differential equation. So, let us say it is y double dot plus E y dot plus d y is equal to 0. I am sure you all know how to solve this. In case you forgotten it is straightforward to check that one of the ways you have learnt in ODE courses is to consider the complementary solution. So, this dot of course as by now you would know it is d dt differentiation with respect to time. So, you take d squared plus ad plus b, but d is some variable and treat it like some you know algebraic variable and solve that equation and find out the roots of that polynomial which in this case is quadratic and then say the roots are m 1 and m 2. So, the solution should be c 1 some c 1 constant times e to the m 1 t plus c 2 times e to the m 2 t very loosely speaking this is something we must have done in your plus 2 level itself right, but here we are going to look at it in a slightly different fashion. So, let of course you will also need y 0 is equal to y naught and y dot 0 is equal to y dot naught just some constants. Of course, you cannot solve a differential equation unless you have given initial conditions otherwise it is just some arbitrary numbers specific solution corresponding to a specific initial condition. So, let y is equal to x 1 y dot is equal to x 2 right and then what do we land up with in this case for instance what is x 1 dot is x 2 and what is x 2 dot x 2 dot is y double dot y double dot that is equal to minus a y dot minus b y, but in terms of x 1 and x 2 what is this? This is equal to minus a x 2 minus b x 1 right. Now, think about it if I want to write this in the following manner I hope you will not object is equal to 0 1 minus b minus a x 1 x 2 is the same system right you agree with of course x 1 0 x 2 0 given by y 0 and y dot 0 actually let me not use the dot notation for constants it is bothering me a bit. So, let us call this p and q. So, this is p and this is q so far so good. So, instead of trying to solve with this differential equation I might as well turn my attention to this it is no difference you agree. One of the things that happens in writing a differential equation like this is you had a second order differential equation now you have reduced it to two first order differential equations, but of course life does not get any easier why primary reason is because after all these are still coupled see the different the solution of our differential equation that is a first order is very simple we know exactly what it looks like. So, for instance if you have x dot sorry x dot plus a x is equal to 0 with some x naught is equal to what should I call it alpha right then what is this you remember what this is this is e to the minus a t yeah that is what your x is let me write it here x t is equal to alpha e to the minus a t you agree can we solve this I mean you can solve it straight away like if you if that bothers you just write dx dt is equal to minus a x therefore dx upon x is equal to minus a dt right which in turn would lead to the natural logarithm of x is equal to minus a t plus some constant is it not a c is the arbitrary constant of integration, but now if you just open it up you will get exactly something like this except that instead of alpha there will be a c here. So, if you plug in the initial condition now that a t is equal to 0. So, this means that so this means that we can write x is equal to e to the minus a t times e to the c right, but then at t is equal to 0 what is it x 0 is equal to alpha is equal to e to the c right. So, in other words I can just say that this is equal to nothing, but alpha e to the minus a t why am I writing it like this because just to show you that why this is appealing this is very easy. So, we might be tempted to wish for a system of equations where we could have purely reduce this to purely two first order differential equations that do not depend on each other and life would be so simple right because whether I write it like this or I had the original differential equation written ultimately it is just a different representation maybe, but in so far as a solution goes it does not seem to ease my job. On the other hand if there was just x 1 dot dependent on x 1 alone why x 2 dot dependent on x 2 alone which would give this a structure of what exactly a diagonal matrix. So, if I had a diagonal matrix sitting over there things would have been a lot easier for me. So, the question then arises this is of course just one form it is not the only form in which you will see second order differential equations in general this is what you will see. So, in general we will have x 1 dot x 2 dot is equal to a 11 a 12 a 21 a 22 x 1 x 2 wave x 1 0 x 2 0 is equal to alpha beta in general we will have this to solve for of course that a matrix that I have just written out over there is a special kind of a matrix it has a special structure by the way this has deep system theoretic implications we will not get there this is one choice of what we call state variables of the system, but anyway let us not get there. So, in general for second order differential equation if you constant coefficient second order differential equation like so if you want to write it up in the most general form this is what you are likely to face. So, no bonus for you that is these two terms that are the cross coupling terms see this term couples the dynamics of x 1 with x 2 if this one there the next one dynamics would have been purely evolving according to the value of x 1 at that instant and it is a first order differential equation yeah. Similarly, in this second case the a 21 is playing foul if this a 21 had somehow managed to vanish then what would I have had x 2 would have been a purely decoupled equation I could have solved it like a first order differential equation. Now, unfortunately none of these is true in the general case. So, how do we go about this then or what would have been a desirable thing. So, think about yourself like you are trying to solve this problem and make it appear like this and you are trying to come up with some conditions some necessary and possible necessary and sufficient conditions. So, that given any such system like this I can always substitute this because after all substitutions are all we do substitute one variable for the other. So, any such substitution that might exist so that I will eventually be able to get this matrix to look like a diagonal matrix and thereafter I can just use this simple tool that I just showed this kind of a solution solve each of them individually in a decoupled fashion and be done with it. So, that is the motivation it is at least you will agree that it is a sensible motivation to have right because it is considerably simplifies the solution of this differential equation. So, and in fact if this does not serve as enough motivation just think about an nth order differential equation where you will have an n cross n matrix here. Of course, here because two dimensions are easy to show so I am taking this, but in general our question will extend to n dimensions right. So, can we or can we not do this and if we can then what is it that will allow us to do it that is the fundamental question right. So, suppose this is the first thing we are doing of course, there is nothing to assume suppose v 1 v 2 is a basis for r 2 such that I mean up until this point it is quite straight forward. So, I am going to call this matrix as a right suppose v 1 and v 2 forms a basis of r 2, but not just any arbitrary basis such that v 1 is equal to lambda v 1 and or rather lambda 1 v 1 and a v 2 is equal to lambda 2 v 2 where lambda 1 and lambda 2 let us say in this case the field is real. So, these are just real numbers suppose somehow someone has told us that this is given right. Now, subject to this can you do that decoupling business. So, I hope this part I can erase now right this part is clear let me say let of t what is this a 2 tuple is equal to let us say x tilde x tilde 1 times v 1 thus x tilde 2 times v 2. In other words this is the same as writing in the language of basis yeah that x t. So, let us call this the basis please ask if this is not clear this is all right agreeable. So, we say that this is an ordered basis now yeah and in terms of the ordered basis I can of course write this as coordinates right. So, then x dot I am going to drop this t sub argument it is understood that it is a variable it is a variable that depends on t because of course differentiating with respect to time. So, x dot, but v 1 and v 2 are constants right because they are just a basis set for r r 2 why should they change with time in this case. So, this is x 1 tilde dot t v 1 plus x 2 tilde dot t v 2 would you agree with me if I write this as v 1 v 2 like. So, this is a matrix now 2 by 2 matrix times x 1 tilde x 2 tilde right dot of course thank you yeah. So, now what do I have here let us call this matrix v yeah. In fact, we can very well write the same thing here x is equal to v x tilde I mean I could have also drawn the same conclusion here that x is equal to v x tilde. So, now what does this mean what is this also equal to from the other on the other hand this is equal to a x where a is a 2 by 2 matrix what is this x v x tilde. So, this is also equal to a v x tilde yeah no problem so far clear. So, what do we have then we have v x tilde dot is equal to a v x tilde as an analogous representation of the same equation is it not of course the initial condition would be different what about the initial condition if you have this as x 0 you can just check that the initial condition will change somewhat. But as far as the dynamical equation is concerned this is the same thing right what can I further say about this is this v invertible it is not is this matrix v invertible or not it is a full column rank of course, because v 1 and v 2 are 2 columns and they are linearly independent because they form a basis. So, it is invertible so why not just invert it I mean I want to get a representation similar to the x dot is equal to a x in terms of the x tilde dot is equal to something times x tilde. So, I can write this as x tilde dot is equal to v inverse a v x tilde is that ring a bell is not that the representation of an operator subject to a change of basis. So, you have changed the variable in terms of a new basis and now you have the operator in terms of the new basis see it is the same connection no in terms of the old basis a was acting on it in terms of the new basis v inverse a v acts on it no difference whatsoever, but is there something interesting about the character of this I wonder look at this. If I now write this as v inverse a v 1 v 2 x tilde which in turn will be v inverse a v 1 a v 2 times x tilde right agreed what is this by my premise what is it lambda 1 v 1 lambda 2 v 2. So, we carve out a little bit of space here that means x tilde dot is equal to v inverse and bear with me a while times lambda 1 lambda 2 0 0 do you agree because this is lambda 1 v 1 this is lambda 2 v 2. So, I am just pulling it out as a diagonal matrix here what is this we are back to v again. So, v inverse times v because this fellow is nothing but v yeah. So, this fellow is nothing but v then v inverse v is identity. So, what I have is let me call this big lambda x tilde and now in terms of this changed basis what I have is a completely decoupled system. So, that means this has seen me through right at the end of the day it is this that has seen me through. Now, if I extend this for sizes greater than 2 do you see any problems in so far as the operations that we have carried out here which is to say if it is a 10th order differential equation or a 50th order differential equation and of course, then you will have the assumption that a v 1 is equal to lambda 1 v 1 tilde a v 50 is equal to lambda 50 v 50 nothing everything carries forward in exactly the same fashion only the size of the problem increases. So, you would have then ended up with a lambda which is a diagonal 50 cross 50 matrix what I am saying is this generalizes I am just showing it for 2 by 2 because it is easy to see it at first glance. But this process that if this assumption holds instead of R 2 to R n with v 1 v 2 tilde v n being a basis for R n such that a v i is equal to lambda i v i for i going through 1 to n. You would have still ended up with the same picture a system of decoupled equations each of them being first order now imagine the benefit this has a 50th order differential equation entails solving for a 50th order polynomial and its roots yeah the complementary solution for a general polynomial unless it has a very special structure beyond the fourth order you will not be able to find any closed form solution you will have to resort to numerical methods. On the other hand if you can make this work of course, I am not saying this is any easier do not go with the misconception that oh this simplifies the solution of polynomials no it still does not address that main problem. It just says that provided these are somehow possible to obtain yeah then you will be able to do this what is so special about this kind of a formulation when a square matrix acts on a vector what does it do to that vector the vector of course, takes linear combinations of the columns of the matrix, but what does the matrix do to the vector yeah scales it and what do you mean by scales now that you have the idea of scaling you should be able to tell me what is scaling what do you mean by scaling how do you quantify that magnitude no great fantastic so we understand so what you can do is you can sort of get an operator which is what a square matrix is to act on a vector to elongate it or shorten it when I say elongate or shorten it means a norm either increases or decreases what else it can also rotate it so in 2d it is easy to visualize those rotations and elongations of course, in 3d also you can visualize but beyond 3d you just have to let your imagination run wild. But the point is these are the things that can do however when you look at this particular vector here there is something very special going on you see it is only a scaling that is happening so these are irrotational rotation in variant directions as if right because when you get this a to act on a vector such as that it results in lambda v or scale times that v it is still going to be in the span of that same vector right it does not leave that span of that vector so it is invariant that span of that vector is a invariant yeah in the language that we have spoken about of course, we have not yet gone into the depths of what is a invariant and all that but when a acts on that vector it is still invariant so these are very special vectors in that sense ok. We will not launch into formal definition maybe in today's lecture but we will carry on with this whatever we have insights we have gleaned about the solution of a second order differential equation through this sort of a substitution and we will see another sort of an application not really application but how we can better represent the solutions of differential equations ok. So up until this point we might have felt that solution of a differential equation means some algebraic expression but there is also sketching of the solution of a differential equation and that is as good as any solution that one might conjure right which is to say that now suppose you want to qualify the kind of solutions that exist right. So, starting from an initial condition does the response of the system decay does the response of the system blow up if it decays along which direction does it decay how does it decay how does it blow up those are interesting questions right. So, of course, algebraically if you write down the solution you might be able to do a bit of analysis but geometrically turns out we can probably do even better. So, for that we will deal with the new basis in terms of the x till days the transform basis and see what we get. So, remember we have two of these numbers lambda 1 and lambda 2 to play around with right. So, now let us call this the x 1 till day axis and this the x 2 till day axis right. So, you already seen that x dot is sorry x till day dot is equal to a x till day. So, in some sense you can figure that if this is the position coordinates of a of an object in two dimensional space and this is the velocity if x 1 and x 2 are just coordinates in the two dimensional space of an object then this is just the velocity the way that position is supposed to alter with time evolve with time yeah that is given by this velocity. So, this is called the field not the field that we have learnt in algebra by the way right at the beginning is a vector field even though it contains vector and field it is got nothing to do with the vector space notion and the field notion it is a vector field of the dynamical system the way in which it evolves. So, do not confuse those notions right. So, now of course, and some x till day 0 given by some I do not know alpha till day and beta till day right ok. Now, suppose lambda 1 is negative and lambda 2 is also negative right and suppose you start from a point here. So, let me use different colors now. So, you start from a point here how do you think you are going to evolve with time? Now, if let me just write this as the lambda right because it has already changed now we know it is diagonal. So, when it is diagonal it is ready made the solution is available what is it? It is just two first order differential equations of the form the first fellow is alpha till day e to the lambda 1 till t the second fellow is beta till day e to the lambda 2 t yeah and lambda 1 and lambda 2 are both negative which means that those exponentials are decaying exponentials. So, as t increases you are starting from here means x 2 till day is 0. So, x 2 till day is 0 means this beta till day is 0. So, how are you going to evolve? Is there any going to be any change to x 2 till day? No right because x 2 till day dot is just lambda 2 x 2 till day and x 2 till day is 0 at every instance. So, it has no velocity. So, no velocity component along the x 2 directions whatsoever no change along the x 2 direction the only changes along x 1 direction, but where to the left or to the right to the left because it is a decaying exponential. So, x 1 must decrease with time. So, starting from here you will as t tends to infinity asymptotically approach this and what is this direction by the way? Is this not the eigenvector corresponding to or the principal vector corresponding to lambda 1 for this matrix? It is a diagonal matrix we have not defined eigenvector yet. So, let me withhold that term for the time b. So, this if you have lambda 1 and lambda 2 and if you want to write this as say x no not x may be z 1 z 2 is equal to lambda 1 z 1 z 2. If you want to solve for this what do you think is z 1 and z 2 and you turn out to be 1 0 which is just this direction. So, this is actually one of those primary directions corresponding to this matrix not the original matrix, but this matrix right. So, you are going to evolve along this direction. What if you had started from here? Again the story is the same in so far as x 2 tilde is concerned right, but what about x 1 tilde now which way right or left and why? Because again that fellow lambda 1 is negative which means the magnitude of that must decrease. So, if you are going to the left then the magnitude of course, it is more negative, but the magnitude is increasing and that cannot happen. As t increases the magnitude must decrease whether it is positive quantity or negative quantity the absolute value must decrease. So, this starting from any point here similarly this is how you will evolve. Let us take something along this and along this. What do you think would happen? I put it to you that you will be moving along this and this direction is it not? Yes? Same argument because both of them are negative after all. So, no change in the argument no, but now the interesting case I mean points on these principal axis it is very straightforward that way too. Suppose you take an arbitrary point here, how are you going to evolve? Where are you going to go? First of all you have to eventually end up at 0 because both those exponentials are decaying, but there are so many different ways no you can go like this I do not know maybe not this because this is just crisscrosses. There is a nice result which says that such ODEs will have unique solutions. So, you cannot have the solution crisscrossing on onto itself right again something we are not deriving, but again I ask you to believe me on that. But nonetheless you understand that it is a meaningful question to ask right like how will this trajectory evolve starting from a point like this? If it is too difficult just think of this like the x and y coordinates of an object of a point mass whose dynamics is given by this then you will immediately be able to see this is the path traced out by it right. Now starting from here how do you think we should go? Now we need a little more information to not to put to find a point on it, but we also need the information about see both of these are smaller smaller than 0. But now we will have to say that which one is more negative? Suppose I am going to draw for this case maybe leave the other to you as an exercise to figure out and it is going to be you can probably even guess right away once I have drawn for this condition. So, suppose this is the case then what will happen what do you think how and why? So, at this point your motion along the x 1 tilde direction is governed by what yeah by lambda 1 that rate is governed by lambda 1 right and your motion along this direction is governed by lambda 2 which one is decaying faster? Lambda 2 because it is more negative it is decaying faster. So, how will you approach this more decay in the x 2 direction? Because if this is a higher thing then by the parallel groom law this is how you should head and then you know piecewise you can string together when ultimately you reach very close to this you still have quite some way to go here right is this argument clear? With that being the case you can now easily infer starting from here you would be approaching like so maybe starting from here you would be approaching like so starting from here you would be approaching like so same things hold here just check that from different initial conditions you will be approaching like this and by the same token starting from here and here this is what we call a phase portrait okay and this is as good as the solution of a differential equation if you can sketch this all right now going back to the more fundamental and important questions. So, of course try out the case where lambda 1 is less than lambda 2 is less than 0 that I leave to you as an exercise. So, this is what I have drawn here and this is what I am leaving to you as an exercise I mean you can probably just write away guess don't give me the answer even if you can let some of your friends try right. So, this is the solution okay next we are not done yet because this is some artificial variables that we have created to ease our understanding but if I now want to sketch this same thing in the original state space of x1 and x2 how would it look yes yeah I mean how we can say that you can make this stick like this you get it yeah yeah because it's the same story. So, 90 degrees is very special and thank you for pointing that out because that's exactly what we are going to now talk about you see here if you look at this matrix lambda and you look at its corresponding v1 and v2 they are actually orthogonal. So, what you have in this special case is a basis that is not just obviously linearly independent but also an orthogonal basis you had an original a in which case you looked for this v1 and v2 those v1 and v2 was just a basis regular basis no like nothing imposed on them but now you have gone to this diagonal form and now if you look for those similar kind of vectors in this diagonal form that satisfy lambda v is equal to like big lambda v is equal to small lambda 1 v1 or big lambda v1 is equal to small lambda 1 v1 big lambda v2 is equal to small lambda v2 they turn out to be just e1 the first principle basis yeah and the second one. So, 1 0 0 1 and they are orthogonal. So, when you have gotten a matrix down to its diagonal form these vectors that you are going to generate by satisfying that condition which I have just erased here are going to be orthogonal and which makes our sketch so easy but now I am putting it to you suppose I okay let me use the white color the same thing now I want to go back to my original state variables because see if you for example take this x1 and x2 to be to voltages and currents and this v1 and v2 is just some abstract vectors that I have chosen which might not have any physical significance maybe you are taking the combination of a voltage plus a current which has no physical significance it is not even physical variable you might have plotted all of this but I am not too happy because I am not seeing the connection between a voltage and the current this is some voltage plus some alpha times some current this is some voltage plus some rubbish some gamma times some current yeah it is a mathematical solution true but from an engineering perspective it has no physical significance. So, I want to actually see how in terms of actual physical variables which are probably x1 and x2 not x1 tilde and x2 tilde how these objects would evolve. So, what is the idea how would we expand our imagination in that case yeah that is an important question recall we still had those two vectors in this domain maybe I should call them v1 tilde and v2 tilde here so that in the original domain here where I had x dot is equal to ax and a v1 is equal to lambda 1 v1 a v2 is equal to lambda 2 v2. So, remember this v1 and v2 were still there very much sitting inside yeah suppose this is the span of v1 right and now v1 and v2 have no obligation to be orthogonal they can be just any two vectors that are not in the same span and they will be linearly independent because it is 2D after all the only way they can be dependent is if one of them is 0 or one of them lies in the span of the other. So, any two non overlapping fellows so this is v1 the span thereof and this is v2 the span thereof now in this I want to be able to draw figure similar to this and then I would say I have a solution in terms I can solve for the current and voltage individually of course algebraically you can back substitute and solve it also but I want a geometric picture we are not going doing any algebra here we just sketching based on our geometric understanding and here the understanding is as follows what is it every point that you pick out here has an image where remember that substitution what did we say x written in terms of the basis where b is v1 v2 is what that is exactly x1 till day and x2 till day is it not remember just maybe flips through the pages and you will see this what does this mean whenever you are choosing any point in the first quadrant of the x1 till day x2 till day space that means these two fellows are positive. So, you are taking x1 till day times v1 which is a positive extension along this direction plus x2 till day times v2 which is a positive around this so it must be a point in this cone within this cone somewhere so a point here gets mapped to a point here. So, these points here get mapped to points here similarly these points here you can readily verify now what is it x2 till day is positive x1 till day is negative. So, x2 till day is positive that means this is still on this side and x1 till day is negative so this is on this side right. So, where what are we talking about sorry what is this this where x2 till day is positive right. So, x2 till day is positive means we are still dealing with the resultant of a vector along this direction and a vector along this direction. So, I have probably flipped the order somewhere. So, anyway so that is going to be a point somewhere the resultant of those two will be somewhere over here. So, I have just flipped the order so do not mind but you see you can always work it out that is the point I am making right. So, you see even though this does not split up the space into four cones that are right angled cones which is very nice here, but this still splits up the whole space into region 1, region 2, region 3 and region 4 where this is region 1, this is region 2, this is region 3 and this is region 4. Unless I have messed up I think that is how the order is right yeah that is x2 positive x1 negative x1 till day negative. So, x1 till day negative means it is v1 folded back yeah I think that is what it is right. So, now what do you think is it going to be the main part the trajectories how are they going to be plotted the argument is still the same because if you are going along this at a point here which vector is it going to approach fast and towards which vector is it going to approach slower yeah because remember this is the condition we are checking this this is what I have left as exercise for you. So, this is the condition I am checking then the decay along v2 is very fast. So, for any point here what are the directions of the velocities this is my direction the direction parallel to this is going at a rate e to the lambda 1 t and a direction parallel to this it is going at a rate e to the lambda 2 t agreed agreed. Now, if that is the case then which one is greater e to the lambda 2 t that component is much greater no. So, this is going to be much greater. So, the resultant is going to be like this. So, eventually you are going to kill off which component faster which one are you going to approach as time proceeds it is exactly the same argument again right. So, the component the decay along lambda 2 is fast. So, eventually you are going to approach as t tends to infinity you are going to approach along this are you not sorry I should almost merge it here like this yeah. So, here if you start for instance you are going to merge along this for instance here like this yeah if you start from here again you are going to merge along this and like so like so if you start from here again yeah it is just v 1 is it not something like this right do you see some pattern here see this one if you if you just imagine some sort of a figure it is just a distorted version it is just squeezed in, but the fundamental character remains the same right this is just orthogonal. So, equally stretched out and this four cones are all right angled cones here, but here two of those cones are really acute cones and two of those are really obtuse cones the obtuse cone gets really stretched, but eventually you end up from any point eventually converging on to the axis which is sort of the principal axis that is along this direction. So, it is a distortion right. So, we will not dwell on this for too long in the next lecture we shall just quickly wrap up with this for different cases which is when both of them are positive I will not give the such detailed explanation. So, make sure that you understand this and when one is positive and one is negative by the way this is called a stable node when one of them is negative and one of them is positive it is called a saddle point and when both of them are positive it is called an unstable node. So, those are the three kinds of pictures or phase portraits that you will obtain when the lambdas are real. So, the next lecture we will quickly cover that part and then move into other interesting topics on Eigenvalues and Eigenvectors starting with the definition of those terms. This is just some sort of an application of how this notion carries forward for dynamical systems. Thank you.