 Morning, welcome back. So, in the last lecture, we have seen that every 2 by 2 system is linearly equivalent to one of the three types that depends on the eigenvalues of the matrix. If the system matrix has two real and distinct eigenvalues, then the given matrix can be diagonalized and if the matrix is has a real, but double eigenvalue, then it goes to a form of the type lambda 1 0 lambda and the third type corresponds to the complex eigenvalues. And we also had the detailed analysis of the type 1 where the eigenvalues are real and distinct and the system corresponds to what is called the equilibrium point is what is called as naught. And if the eigenvalues lambda mu, the product of the eigenvalues is positive and the product of the eigenvalues lambda mu is negative, then it is called as saddle point equilibrium. And that is an unstable equilibrium. When lambda mu greater than 0 and both the eigenvalues are positive, then the equilibrium point is unstable and if both lambda and mu are negative, then the corresponding equilibrium point is stable. So, that is the type 1 analysis completely. Now, we will quickly go to the type 2 and type 3. So, we will go to the type 2 type. So, type 2. So, the type 2 there are two cases when b 1 is of the form b 1 is equal to lambda 0 0 lambda. This is the case in the type 2. There are two cases namely case 1 case 1 where the eigenvalue is a double eigenvalue, but it has two independent eigenvectors. In this case and of course, lambda not equal to 0, this situation is absolutely no difference. It is exactly like type 1 where you can write down lambda equal to 1 and it is like a node. So, the equilibrium point is a node. Exactly like in the previous here previous class of the type 1 and absolutely no different. It is stable if lambda positive lambda negative. It is unstable if lambda positive. The solution can easily be written as in this case x 1 t is equal to x 0 1 into e power lambda t and x 2 t is equal to x 0 2 e power lambda t same lambda. So, this if you eliminate t you will get x 1 is equal to a constant in the x 2. So, it is nothing but straight lines all the time. So, if you have the phase for trade here you have the origin as an equilibrium point x 1 equal to x 2 is nothing but a straight line. So, if you start any point here and if this is the situation when lambda negative it will be stable. So, it will move along that and it will this is the. So, any point you start it will be completely like this wherever you start the straight line. So, the entire trajectories are given like that it will go. So, that is a phase for trade for lambda less than 0. If lambda greater than 0 I will draw it in the same thing if lambda greater than 0 it is the same thing if you start with, but then it will move away from that. So, this is the for t negative part from here if you start. So, if you start anywhere it will move along this straight line this is the point t negative part. So, it will move along this trailer. So, trajectories are straight lines. So, if you start from here again if you start from this is the starting point and starting point it will go along this trailer. So, you will have unstability this is a phase portrait phase portrait of two systems is not just one for lambda less than 0 you have the complete. So, phase portraits of for lambda greater than 0 you have your unstable thing that is nothing of completely r 2 and for lambda less than 0 it is a completely stable there is no unstability or instability. So, you have the complete e u is equal to the stable subspace and unstable subspace are given like that. So, there is no thing. Now, we will go to the case 2 case 2 in the case 2 if you do again it behaves exactly like the node case. So, there were not real difference between type 2 and type 1 when the real distinct taken values. So, only thing in the type 1 when lambda mu less than 0 it is a saddle point equilibrium that is only difference otherwise it is a node. So, for this case so the b 2 is of this form lambda 1 0 lambda. So, the corresponding system is x 1 dot is equal to lambda x 1 plus x 2 and the x 2 dot is equal to lambda x 2. So, the lambda for x 2 is an independent thing. So, you can solve for you can solve this system separately if you prefer because if x 2 can be solved immediately because it is a decoupled part. Once you find the x 2 you can put it here for x 1 and that is a non homogeneous system for x 1 otherwise you can immediately compute. So, here is an exercise for you to compute now this is an exercise one should do it immediately you for this matrix for diagonal matrix you already know if it is a diagonal entries lambda 1 etcetera lambda n e power the whole thing is the computation of exponential. So, it is a diagonal matrix the diagonal entries will be e power lambda, but how does it look like if for a b 2. So, that is where the exercise. So, you compute your b 2 square b 2 q etcetera and then you compute e power b 2 this computation is easy tell you we are doing this because the computation is not easy for us and also for general matrix does not reveal the trajectory. So, this you can compute to see that e power b 2 is e power lambda 1 1 0 what is this. And, also compute because we are interested in writing the solution you also compute t b 2 t square b 2 square t q b 2 cube etcetera and write down e power t b 2 this is exercise is not hard you just have to do it familiar you get e power t lambda into the matrix will be 1 here t will be here. So, rho will be 1 will be here. So, you can write down now the solution immediately because you can write down the solution. So, your solution will be. So, you can do it either computing e power b 2 or you can directly solve for this system. So, your solution x t will for this system e power t b 2 into x 2. So, if you decouple the difference. So, you will get x 1 t is equal to immediately you can decouple x naught x 0 1. So, you will write down your solution x 0 1 you can do the computation x 0 to the initial values of t e power t lambda x 2 t is equal to x 0 1. Because it is an independent of that. So, x 1 would not come there x 0 2 into e power t lambda. So, you see. So, this will go to plus or minus infinity if lambda positive as t tends to infinity this will go to 0 if lambda less than 0. This is the same case for this also this will go to plus or minus infinity if lambda positive this will go to 0. So, you will have stability if lambda. So, it is again the 0 is called a node it is exactly like that called a node. The only difference is that because of the appearance of this term when you plot the trajectory in the neighborhood it may have a different behavior, but then there is a t here this will go to infinity the effect of x 0 1 will be wiped out. So, you will see so eventually it will be behave like this term because of the t. So, if you plot the solution here. So, for in a typical case if you face portrait if you do if you start here in the neighborhood there will be a twist because of the term like this at all. So, the exercise again is to plot this curves in the properly. So, you may if you start with the trajectory for lambda less than 0 may come if you start with initially there will be some twist it may behave like that it will go to the origin if you start here. So, from here if you start it may. So, it will go to 0 if for the lambda case it will be something like that there may be some trajectory there may be some trajectory difference of twisting because of the neighborhood of the origin there will be, but eventually it will go to the origin. On the other hand for lambda positive if you want to see the lambda positive case the trajectory will be the same, but then it may if you start from here the trajectory may go like this the same behavior s will be there if you studied only thing it will go to infinity it may it depends on the initial values. So, in the neighborhood there may be a motion difference, but eventually either it will go to 0 along a curve given by this x 1 t and x 2 t it is given by the parameter representation x 1 t x 2 t it will go to 0 as lambda less than 0 and it will go to infinity plus infinity or minus infinity from which quadrant you are starting with and the behavior of that one. So, basically it depends on the this here the sign of that. So, it is a case of node. So, the type 2 is exactly the one case of type 1 in the node situation. So, again the equilibrium point in all these situations we call a node and there is a node is stable if lambda positive negative and it is unstable situation unstable if lambda positive. So, that completes the situation of type 2. You have the your type 3 you have your a Eigen values lambda is equal to a plus i b lambda bar is equal to a minus i b and your matrix b 4 is equal to a minus b b a this is the key. So, your exercise will be to compute for this special type of the general matrix it is difficult the matrices we came upon is easy to compute. So, I want you to compute as an exercise e power b 4 that is what our always the job will be e power a and computation will be cos of b minus sin of b it will be sin of b cos of b this is your e power b, but we are interested in e power t b 4. So, you have to compute this one if you compute this one you will get e power t a and matrix will be cos power b t minus sin b t sin b t and cos b t. You can directly compute b 4 b 4 square b 4 cube b etcetera or t b 4 etcetera another way to compute probably this is then may be this is easy to compute you can just try if lambda is equal to a plus i b you can actually see that a power k you have to compute no a not a plus i b you will have b 4 power k will be please check this the validity of this one it will be real part of lambda power k minus imaginary part of lambda power k imaginary part of lambda power k real part of lambda power k that is why you get it cos b if you add it b 4 power of factorial you exactly get cos b t sin b t etcetera you can do it that b 4 of the you have to work with the t b 4. So, correspondingly there will be t here. So, you can do these things exactly yeah. So, with that you can write down your solution now. So, what is the system corresponding to this one corresponding system in y system is y 1 dot is equal to a y 1 minus b y 2 and y 2 dot is equal to b y 1 plus a y 2 and if you write your solution for the system solution y t is equal to e power t b 4 y naught. So, we already come computer e power e power t b 4. So, it will be e power t a into cos b t minus sin b t sin b t cos b t acting upon some y naught. So, you have a representation of the solution. We want to see the phase portrait the aim is to understand the phase portrait. So, let us denote for the time being denote c is this matrix c is the matrix cos b t minus sin b t sin b t cos b t. So, this is the exercise part therefore, your solution y t is in a short form we are at e power t a c y naught where c is this this is a temporary notation. So, what here to understand the phase portrait that is what you have to understand now as t tends to infinity look at these terms here depending on the sign of a depending on the sign of a e power t a will go to plus infinity or minus infinity if a is negative t power t a will go to 0 if a is positive t power a will go to infinity if a is on the other hand what will be this is doing this term cos and sin are periodic with period 2 y cos t and sin t a. So, this shows the second term here when act on y naught shows some rotation because as t tends to infinity these terms this action of this c c y naught if you start with a point y naught it will rotate around the origin that is what c will do. So, c on the action of c on y naught will try to rotate around the origin on the other hand the term t power t a if a not equal to 0 and that will either that is typically the amplitude if a is positive e power t a will go to infinity. So, it will start rotating, but rotation the radius of the rotation will start increasing if a is positive because that is the amplitude if a is negative that will it will go to 0 the amplitude. So, the e power t a will give you the amplitude of the rotation this will give you the periodic rotations around that point. So, we have to distinguish the case two cases when a not equal to 0 and a equal to 0 because when a equal to 0 e power t a will vanish and there is only a pure rotation there is no amplitude change leading to what is called the periodic solution you will would be studying these things in the non-linear analysis more looking for periodic solutions of the trajectory. So, we will consider in this case two cases case with a not equal to 0 of course, recall that of course, b is always not equal to 0 in type 3 because if b equal to 0 then there is no complex part there is no imaginary part the Eigen values are just real and the real case we have already studied. So, of course, b is always not equal to 0 in type 3 because we are in this situation of complex Eigen value. So, we will start with a is a not equal to 0 this leads to a not equal to 0 complex Eigen value. So, we will start with a is a not equal to 0 this leads to four cases according to the sign of cases according to the according to the sign of of a and b. So, what are the four possibilities coming into picture? So, let us try to understand one by one the case one. So, we will have two cases first we brought let us plot the case when a positive when a is positive the amplitudes are increasing. So, if you are starting from any point here it will rotate, but then rotation can take place either it will rotate in such a way that the amplitude should become bigger and bigger in both cases, but the only thing is that whether it will rotate clockwise or anticlockwise. So, there are two options for a positive the rotation can take place in the clockwise direction and the rotation can take place in the anticlockwise case. So, what the exercise I am going to suggest here you try to take if you want to say whether clockwise or anticlockwise what is the sign you have to choose you start with some particular values of b and also particular values of a. So, consider two specific situation choose some b say equal to 1 and choose a equal to 1 and then you take b equal to minus 1 equal to 1 and all that possibilities you try to see that rotation how it take place to see that one specific case if you want to see. So, let us take the case for I want to know that the rotation amplitude is going to infinity. So, what will happen is that if you think that one go. So, one possibility is that it will take an anticlockwise direction and the amplitude will go that is the case. So, your arrow will be something like that the other situation if you take the second amplitude is increasing the clockwise it can go you see it is going to infinity. So, this is your situation with y naught this is the y naught here the only thing is that one will be for b negative one will be for b positive the best way to check is that you consider a specific case for example, with b positive what will happen to that if you want to check here say choose b equal to 1. So, your matrix is cos t minus sin t sin t cos t this is the way you have to look at it and look at see the vector which is starting from 1 0. So, I am choosing a vector from here I am choosing a vector from here or you can see that vector from there it is choosing and what will happen this one this will be suppose I take t equal to 0 that is a situation when t equal to initial point is at the origin. So, when t equal to 0 cos 0 is equal to 1 sin 0 is equal to 0 0. So, it will be 1 0 0 1. So, it will go to 1 0 only at t equal to 0 where it will happen suppose I am taking a rotating pi by 2 at taking t equal to pi by 2 at a later time. When t equal to pi by 2 cos pi by 2 is equal to 0 now sin pi by 2 is equal to minus 1 and sin pi by 2 is equal to minus 1 which I want you to see that that is equal to and this is 1 0 it will move to 0 1 if you compute it. So, what will happen is that if I start from here I will have my situations here this is with if I rotate here next point will be it will take this point y naught if I start y naught from here it will go along this it increases. So, this is the situation with b positive clockwise and this is the situation b negative. Now, this is the case with a positive now you can verify when two more cases will come what are the two cases both these cases are with a negative in this case again the same thing you will get a clockwise rotation you will get a clockwise rotation if you start, but then the amplitudes are reducing. So, it should go towards the origin. So, it will come here in the using. So, this is the situation with b positive then if you start from here if you start from here see it will go to the origin the spirally. So, this is a negative with b negative this is a positive with a negative with b positive. So, to get this you see of course, we are taking a minus b b a that is a matrix we consider certain places you may take this also may be written in the form of a b minus b a, but you have to check the sign properly look for thing. So, the exercise is the best exercise is that take some particular values of a and b with a different all the four combinations and try to focus here. So, and this situation is called focus. So, we will have this is called the focus it is a stable focus in all the situation all these four situation we call the equilibrium point a focus and stable if a negative. So, it is depending on the real part of the Eigen value the stability always depend on the real part of the Eigen value not about the complex part the sign of complex part determine the orientation of the trajectory. So, you have a two focus is unstable in the first situation first two figures and it is stable in the next two figures. So, we have left with one more case with a equal to 0 case now the last case which we want is case with a equal to 0. In this case the Eigen values are purely imaginary and your solution y t is nothing, but you see cos b t is nothing but minus sin b t sin b t cos b t it is a pure rotation it will rotate pure rotation see again. So, there it is always come back to if you have the trajectory. So, if you are trying to do a phase portrait here and you have a equilibrium point it will only thing is that it will rotate like this the other situation which you get is that it will rotate like this that is only two possibility again it you have to see that which direction it will see this is the clockwise direction with earlier case we have already seen that. So, you have this situation. So, you have you see this is a b case positive case you have an clockwise and b negative you have only clockwise. So, you have so you have this is the case with the b positive and this is the case b negative such a point here 0 is called a center called a center. So, with a non 0 it is a focus in the complex situation when a equal to 0 it is a center the equilibrium point is called a center completely. So, that giving in 2 by 2 systems we have a complete analysis of this one. So, let me give a fine definition finally consolidating all the results we will write the definition for a 2 by 2 system for a 2 by 2 system any system one system one what is the one system let me recall x dot is equal to A x the equilibrium point the equilibrium point 0 is called one saddle point if lambda mu negative node here that the moment lambda mu is negative it cannot be complex second values. So, it is a automatically it tells you that is second value to a node if lambda mu real here you have to say it is real and lambda mu positive. Lambda mu positive because if lambda mu lambda mu even in the complex case will be real even if lambda and mu are complex because lambda will be a plus i b mu will be a minus i b and hence lambda mu will be a square plus b square. So, it is always positive and it is unstable if lambda positive mu positive stable if lambda negative both the eigen values are negative and third situation the equilibrium point is a focus called a focus if lambda mu complex with real part of lambda both real part of lambda is not equal to 0 because real part of lambda is not equal to 0 and that is also a condition stable if real part of lambda is real part of lambda is negative unstable if real part of lambda is positive the last case for a center if the same thing lambda mu complex with real part of lambda equal to 0. So, you have a in other words purely imaginary that is pure imaginary. So, we have a complete analysis of the 2 by 2 system there is an interesting bifurcation diagram will end this 2 by 2 part before going to high dimension there is a bifurcation diagram nice bifurcation diagram. So, you can get a picture this bifurcation diagram is in the bifurcation diagram this is in determinant trace space determinant trace plate what is what do you mean by that determinant trace plane given a thing you given a matrix you can write determinant of a is nothing, but the product of eigen value let me call it this is to delta by definition and then you call it alpha the trace of a that is nothing, but the sum of eigen values then you know that clearly because this is the product and this is the sum and it solves a second order algebraic equation. So, this would be in the coefficient of the power term. So, the eigen values satisfy the eigen values satisfy the quadratic equation right quadratic equation z square minus alpha z that is the sum of the eigen because it is a roots. So, it is a coefficient of that plus delta equal to 0. So, you can write your lambda mu as solve that equation lambda mu the eigen values can be written in terms of the determinant and the trace that is equal to alpha plus or minus square root of alpha square minus 4 delta by 2 you see. So, you can write your alpha is one is the alpha plus square root of alpha minus 4 delta by 2 and the other one is alpha minus square root of alpha square minus. So, I am going to draw the diagram in the plane determinant by this is alpha that is nothing, but the trace and this is delta delta is equal to lambda mu and this is equal to lambda plus 1. So, this is nothing, but alpha is equal to lambda plus mu. So, what is this? So, let us consider all the cases. So, below this domain below the alpha x is delta. So, this is the region delta less than 0 delta is equal to lambda mu less than 0. So, what is this region? This region is this region the complete region complete negative region on that you have all 4 points for lambda mu satisfy it belongs to this region that means delta less than 0. So, you have your saddle point here. So, you have your saddle you have your saddle point this is the saddle point this region saddle point. Now, let us look at the situation here when this Eigen values are real Eigen value it is real. So, you have lambda mu real if alpha square minus 4 delta greater than or equal to 0. So, you have to consider the equation in complex if alpha square minus 4 delta less than 0. So, in the alpha delta plane alpha square minus 4 delta equal to 0 determines a parabola. So, you have to draw the parabola here now you have to draw the parabola here. So, you have your parabola here and this is nothing but the region alpha square minus 4 delta equal to 0 and what it shows that. So, you have a different thing this is the region alpha square minus 4 delta positive this is also the region alpha square minus 4 delta positive here is alpha square minus 4 delta negative and here alpha square minus 4 delta negative. So, that gives now you will consider this case again on that case when alpha square minus 4 delta greater than or equal to 0 1 Eigen value here with the plus sign here will remain to be positive. So, 1 of the Eigen values minus that will be negative because alpha square. So, if you look at here alpha square minus 4 delta positive this region if you look at all this region corresponding to this one alpha square minus 4 delta positive you will have the stable focus unstable in this case it will be unstable unstable here is a region with node. So, you have unstable node because it is a situation where you have real Eigen value. So, you have only node here also this is a situation it is a node thing, but then you have the other thing and you will have your stable Eigen value here. So, you will have this stable here stable node you see. So, this is the situation alpha square minus both are nodes, but here you look at it here it is alpha is equal to with alpha positive and this is the region with this part is the region with this part is the region with alpha negative. So, you will get your Eigen values in this region both Eigen values will be negative in this region we will look at it. Similarly, if you look at here you will get unstable focus unstable focus and you here you will get a stable focus. So, you have to understand that this is a region this entire region is where delta is equal to lambda mu positive. So, you will have a region here delta lambda mu positive with lambda plus mu positive that gives you both lambda and mu are positive and here you will get this stability. What about this here this region this line on that line you will see that the real part is 0 the real part is 0. So, that is the region where you have center you see. So, this is the region. So, this gives you in the delta alpha plane a complete analysis of the you have your complete analysis of the. So, this portion the first quadrant first quadrant is the unstable situation quadrant of course, the third and fourth are already unstable because of this added point equilibrium the first quadrant is unstable unstable and this is called end is called the end is called the source and second quadrant is stable called part this is called the sink see everything will sink there. So, this gives you a complete you do not have to think in a determinant press plane you can completely understand where your stability and instability is happening where it is added point where it is not where it is focus where it is thing. So, that gives you the a complete picture of course, we are in an easy situation of 2 by 2 and hence you have the complete picture such a picture in general is much more harder when you go to the higher dimensional situation. So, we will now try to give some interesting 1 or 2 examples and before going to higher dimensional situation. So, we will at least give 2 examples. So, here is an example 1 we will have more examples later example 1 example is a situation where your eigen values previous thing there is one more thing what you where is the picture. This is the situation this case let me have a different color this is the case where you have the degeneracy you see this is the situation where determinant is 0 which is a kind of degenerate situation is a degenerate you will have the degenerate case that is a region that is what we skipped. So, that is a we will see an example of the next one. So, consider an example with a matrix a is equal to 0 0 0 minus 2. So, what are the eigen values eigen values are lambda equal to 0 and mu equal to minus 2. So, we want to determine of course, x equal to 0 for any linear system the origin is an equilibrium point, but if a is invertible origin is the only the equilibrium point because the equilibrium points are given by a x equal to 0 set of all x such that a x equal to 0. So, if invertible only x equal to 0 is the solution to that one, but here is a case determinant of a not equal to 0 this is a case determinant of a equal to 0. So, you have to look for what are all a x equal to 0 you want to find all the equilibrium point a x equal to 0 immediately will give you x the second component this is x 1 x 2 the second component is 0 that is all you give first component it does not give you anything because the equation say x equal to 0 is equivalent to 0 x 1 plus 0 x 2 is equal to 0 second one is 0 x 1 minus 2 x 2 equal to 0. So, that show x 2 equal to 0 that mean every point on the x axis the x 1 axis every point on the x 1 axis is an equilibrium point it does not put x equal to 0 does not put any conditions on the first component of x that implies any point every point every point on the x 1 axis is an equilibrium point you see. So, equilibrium is a equilibrium point so now I want to solve this equation. So, what are the equations you do not have a the equation the first equation the is equal to x 1 dot is equal to 0 because say x equal to 0 implies x equal to 0 x 2 dot is equal to minus 2 x 2. So, if you solve this equation you will get x 1 dot is x 1 t is a constant that nothing, but a constant is the initial point and x 2 t is equal to x 0 2 into e power minus 2 t you see. So, what does it shows that now if you are trying to find a this is nothing, but an equation of a line, but it is an equation of a perpendicular line. So, that is why it cannot be written as a function of x 1 variable and anyway x 1 this is x 2. So, what I am trying to say is that every point on this is an equilibrium point all the points are equilibrium point not just the origin all points are equilibrium points and what does it say that. So, if I start a point from here x 0 what does this say that if the first component this x 1 trajectory says that the first component will not change. So, this will be your x 0 1 this will be your x 0 2. So, if you start from here the first component will not change and the second component says that it will go to the same x 0 it will go to the 0 this goes to 0 as t tends to infinity. So, the second component goes to 0 and x component will not change it will remain if it does not change it has to come along this. So, it will move along this. So, whenever you start here the trajectory will move here as t tends to infinity you start from here it will move to that anywhere you start it will go towards that because x 2 component. So, all your trajectories are given like this all whatever your trajectory you start it will move towards that. So, this is the complete phase portrait of this system phase portrait. So, in other words starting from. So, you are starting from x 0 1 x 0 2 at time t x 0 2 at time t x 0 2 at time equal to 0 the trajectory moves to towards moves towards only at infinity moves towards x 0 1 origin along the perpendicular line along the perpendicular line you see the behavior difference in that one. So, we will go to another example how the linear transformation linear equivalence example will change. So, you will start with a system which is not in its normal form. So, you consider a system consider a system let me write down x 1 dot is equal to. So, part of it is an exercise also minus equal to minus x 1 minus 3 x 2 and x 2 dot is equal to your 2 x 2. So, your matrix A is nothing but minus 1 minus 3 0 2 this is your A here is an exercise for you. So, let me put it in a different color the exercise show that the A Eigen values are clear A Eigen values immediately know A Eigen values are lambda equal to minus 1 mu is equal to 2 with A Eigen vectors find the A Eigen vectors with A Eigen vectors with A Eigen vectors 1 0 minus 1 1. Therefore, the matrix P which converts to a linear equivalent matrix because these are 2 distinct Eigen values. So, you immediately get as it is a 1 0 minus 1 1 compute P inverse compute P inverse P inverse is nothing but 1 1 0 1 and that will give you your B equal to P inverse of A P that is of the form minus 1 0 0 2 you see and that is a diagonal system as expected because you have 2 distinct Eigen values and hence you will have thing. So, what is a y system therefore, your y system is y dot is equal to A y that is already we have studied it is a standard problem which an saddle point equilibrium now is clear because you have 2 real Eigen values with opposite signs and hence it is a saddle point equilibrium. So, you can write down which we have done already the solutions. So, if you plot this curve here immediately I do not want to go further here to you can do that one. So, if you plot your y this is for the y axis this is your y 1 this is your y 2 and this also says that the first Eigen value the corresponding to the first Eigen value minus 1 and you have a the first component stability the second component stability totally instability, but then it is a saddle point equilibrium you will have this one. So, this is your phase portrait here phase portrait and y 2 go to infinity and y 1 go to 0. So, y 1 go to 0 means your graph will be like this. So, y 1 will go to 0 you see it will go to 0 this will go to 0. So, this is. So, this is where phase portrait of that and it is an invertible matrix. Now, we want to know the solution corresponding to x 1 you can of course, do the analysis because you can write down from here e power a t you can write down a is nothing, but p b p inverse you can write down the solution. And then your solution is you can write down as e power t a is equal to e power p a of x naught is nothing, but p b p inverse of x naught p b p inverse of x naught. So, you choose this p inverse of x naught as y naught. So, this is starting with that p b of y naught and p b you can write down the solution completely immediately. So, the solution is x t is equal to e power t a of x naught is nothing, but p you can completely compute b is the diagonal matrix. So, it will be e power minus t 0 0 e power 2 t of p inverse we already computed. So, this is your solution. So, you have your solution completely you can plot it separately, but then let us try to understand the transformation in this slide itself. So, that we do not need in another way. So, what is your transformation going? So, your transformation y going to x equal to p y this we already discussed. So, you compute that one if you do that transformation how does my axis I told you in the beginning this kind of transformation linear equivalence is the coordinate change. So, I want to know that how does my y 1 coordinate changes. So, I want to first compute p of 1 0 if you compute p of 1 0 you just take p is equal to 1 0 is nothing, but 1 0 itself. So, the y 1 coordinate axis goes to x 1 axis itself, but what I do to my second coordinate axis p of 0 1 the y 2 axis and the y 2 axis you can see that it is minus 1 1 that is what it will go to 0. So, if you plot. So, that if you say if you know try to sketch my in the x 1 x 2 plane the if I sketch the x 1 the y 1 axis go to your x 1 axis. So, I will plot here. So, let me choose a proper color. So, if it is you see it will go to the same color, but then the x 2 axis goes to p 0 1 goes to minus 1 to 1. So, it will go to this axis it that is what the y 1 minus y 2 plane. So, it will go to this plane. So, this is the new coordinates now this is y this is the coordinate y 1 plus y 2 equal to 0 or x 1 plus x 2 equal to 0 you see that is where it goes. So, the y 2 axis actually transform this one. So, now the trajectories do not intersect. So, the trajectory corresponding to this one if you take it will go like this. So, the trajectories will be like this and the corresponding trajectories will be the even the directions will not change it. So, the directions will be like this. So, the trajectory here it will be only so the saddle point equilibrium. So, this will be the trajectories in the. So, that is the trajectories here and this is the trajectories here. So, this will all will be unstable the second component is unstable the first component is stable. So, it is till you get the stability here this will be in this direction you know there is not we are not talking about that. So, this will be here this is direction. So, this is the phase portrait phase portrait of the solution phase portrait of the solution. So, that gives you that is what in the beginning I have told under linear equivalence under this thing the trajectory behavior the equilibrium the stability of the behavior do not change whatever stability you have the you will get the same thing. So, the picture something like will be changed it to transformed into a new coordinate system. So, with this the will stop the complete 2 by 2 part now we will not be able to do such a detailed analysis for the higher dimensional system, but we will appeal to the Jordan decomposition what are the best possible things we can do it in higher dimension. We will begin with one or two examples and possible ideas behind it and then we will see what are the blocks coming in higher dimensional system in the next lecture. Thank you.