In this course we have discussed so far about various aspects of microfluidics and microfluidics is actually like an ocean. So you could have discussed many other topics and essentially I have tried to confine myself to some of the very commonly addressed topics in microfluidics, but there are several other topics which I did not touch upon, for example, acoustofluidics, several other topics of microfluidics which can be of emerging importance. However, it is also important to understand that nowadays with advancement in manufacturing technology or with advancement in fabrication, it has been possible that one can make channels of dimensions significantly below what could be possible over the past few years. So if you can make channels or if you can fabricate channels which are of dimensions smaller than the microfluidic domain then at some point of time you enter into a domain which is called the nanofluidic domain. While it is important to appreciate that sometimes I mean issues in microfluidics and nanofluidics it depends obviously on the length scale and the physics and the time scale and all these things under consideration that in many cases the microfluidics aspects and corresponding nanofluidics aspects are more or less very very similar, but in certain scenarios you have to take additional effects into consideration when you are thinking of nanofluidics. And I just want to give you a broad perspective that when we are thinking of flow through a microfluidic channel which is of dimension of say 10 micron or so then the physics of that is as good as like flow through a channel which may be of the which may have hydraulic diameter of the order of say 100 nanometer. So, although we have come down from the micro to the nanodomain that does not mean that the physics suddenly gets altered, however there is a critical length scale below which certain additional interfacial interactions tend to become important which were not that much important over the microfluidic domain and that is like a sort of a very critical area or a very critical domain of nanofluidics that one would be interested to consider. Keeping this in view we will discuss some important aspects of nanofluidics I mean I can tell you that nanofluidics is again a huge area of importance and nanofluidics can be studied by various techniques including experimental techniques, theoretical techniques, computational techniques like that. I will make no effort in going through a comprehensive review of understanding nanofluidics because this particular course is primarily dedicated towards understanding microfluidics. But because there are certain areas of common interest and overlap and certain effects which we get in micro channels become even more important like for example slip phenomena may become even more important if you come down to the nanofluidic domain. That means that if you have interest in area of microfluidics then with modern advancements in manufacturing you should attempt to probe certain interesting areas of nanofluidics which could not be experimentally probed maybe even 4, 5 years back and it is a highly emerging subject and there are issues which we will try to appreciate that how certain issues in microfluidics will differ from issues in nanofluidics will differ from the corresponding issues in microfluidics and we will try to see that how we can address these issues. So after we discuss nanofluidics we will discuss a little bit about molecular dynamic simulations so which are commonly used to capture the like dynamics over nanoscopic length scales where molecular effects are important. So we have to understand very carefully that not over all length scales ranging in the nanometer domain individual intermolecular interactions become or the discreteness of the molecular entities become that important but when you come down below a threshold dimension that becomes significantly important. So with this little bit of background we enter into the discussion of nanofluidics so the outline of our discussion will be first we will discuss about some motivation which partly I have discussed the contrast with microfluidics the interactions at small scales the nano versions of some celebrated equations. So nano versions this is a very loose terminology that we have used here so by this what we mean is that modified version to take into account some of the nanoscale interactions. So the Lucas-Walshburn equation if you recall that we use the Lucas-Walshburn equation for capillary filling dynamics. So the capillary filling dynamics modification of the capillary filling dynamics equation and the modified Poiseu-Boltzmann equation and then we will discuss some applications of nanofluidics. So nanofluidics mainly concerns the study of behavior manipulation and control of fluids that are confined to structures having characteristic length scales ranging from 1 nanometer to 100 nanometer. But this is a sort of very loose definition the physics entirely changes as you go from 1 nanometer to 100 nanometer. So you cannot discuss the some common physical basis by which you traverse this kind of length range of length scales. Nanofluidics of course the nano originates from the Greek word meaning dwarf and SI prefix for 1 billion that is 10 to the power minus 9 and the fluidics which originates from a combination of pneumatics, hydraulics and electronics on a foundation of fluid mechanics. So you can see that the wide gamut of activities that can be centered around nanofluidics. Now question is why are we interested about nanofluidics? Is nano just a buzzword because of which we are interested to study nanofluidics? It is not exactly like that. With nano size channel we may tend to approach the molecular dimensions. The concept of continuum no longer remains defined at this scales. Standard analytical expressions at micro scale can no longer be applied for nano devices but we have to understand that what we essentially mean is that we can still apply that for nano devices but not for those cases where the individual discreteness of the molecular entities become important. So then we cannot use so that is below a threshold length scale not necessarily 100 nanometer but if you go down to 10 nanometer or even less then those things become significantly important. Massive increase of surface area to volume ratio with decrease of size this we have discussed in one of our very early lectures and that means that surface effects can dominate to such an extent that it can open up newer areas arenas of transport processes. Close to the wall every flow has a particle based nature and nanofluidics plays its role. This is a very important consideration. So if we are interested about length scales which govern the physical behavior close to the wall then that length scale if that length scale is important for the physical problem under consideration then nanofluidics is important although the system length scale may not be nanometer. So that is also something important so it is not necessary to go down to nano dimensions of the channel to observe some effects of nanofluidics. Nanoflows versus micro flows although governed by a common physics as one scales down from micro to nano the well understood laws of continuum mechanics remain no longer valid. I mean again I am giving you a warning that this is not a hard and fast comment that I want to make. The laws of continuum mechanics may still be valid in the nano domain it depends on what are the length scales and what are the time scales etc that you are addressing. So for example for pressure driven flow through a 100 nanometer channel I mean you can safely pressure driven flow of water through a 100 nanometer channel you can safely use some of the well known equations governed by continuum mechanics with certain modifications but those modifications will tend to be need to be more drastic as you come down to may be 10 nanometer 1 nanometer like that because then actually in a true sense if you come down to 1 nanometer the continuum hypothesis does not work anymore. With the interfacial and the bulk scale having the same order of magnitude laws of differential calculus may be unable to capture the gradients in properties and at this scales one requires a particle based investigation technique like the direct simulation Monte Carlo or molecular dynamic simulations. Some applications nature has been employing nanotechnology over millions of years and effectively mimicking it can create wonders. So this is something very very important in nature you know there are certain observations which if you can make and you try to mimic that observation in the form of an engineering device over nanoscopic scales then actually that can give rise to a very high level of innovation and there are several research groups which actually attempt to do this. I mean there are dedicated researchers who will just observe nature by observing nature they will try to identify some mechanism and then engineers will take that up and try to make devices which mimic that mechanism to the extent possible and this kind of mimicking nature is also known as biomimetics towards the end of this course I will give you one or two examples of some work from our own research group with emphasis on biomimetics. Now in nanoscale the immensely enhanced surface area leads to dominance of interfacial effects greater area available for chemical reactions and potential of large power density and you I mean in one of my lectures towards the end of this course I will show you that how you can use nanofluidics for making a very high power density energy conversion device. So that is basically using the concept of streaming potential which we have already discussed in the context of microfluidics. So in nanoscale you can do more efficient mixing and separation with applications in desalination, DNA hybridization and so many other areas. Of course you have to realize that when we are thinking about mixing in nanoscale you use a different strategy to create vortices in nanoscale you can pattern the surface with nanostructures and therefore patterning of surfaces with small scale structures in the nanodomain is one of the very hot areas of research these days because you can do wonders with this technology in the fluid mechanics you can create nice patterns and vortices in the flow structure to get good mixing. So nanotechnology has allowed one to use to combine fluid mechanics with that and achieve good mixing over nanoscopic scales. Now we will discuss some important physical effects in the nanoscale domain in the context of nanofluidics. Hydrodynamic interactions what we discussed first it is not just important in the nanodomain but it is important in general in the context of fluid mechanics. But in the nanodomain hydrodynamic interactions may play a big role what is hydrodynamic interaction I will basically not go through the detailed mathematical treatment of this because I mean this is very involved and I will just try to give you some glimpse of what it is so that if you develop interest you can read by yourself. Presence or transport of a finite size solid body creates disturbance in the surrounding fluid the surrounding fluid may be at rest or in motion such disturbance gets transmitted to affect the motion of the bodies that are in vicinity of the original body. This effect is known as hydrodynamic interaction why nanofluidics it is very critical because hydrodynamic interaction gets largely altered in presence of confinement. So make the confinement smaller and smaller this becomes more and more non-trivial bulk hydrodynamic interactions that is without confinement effect. So bulk hydrodynamic interactions are commonly accounted by considering following simplifying assumptions. The disturbance created by the solid body is assumed to be a point force this effectively means that the solid bodies can be treated as point sources of drag in the solvent acting at a point typically the center of a spherical body. This disturbance or effectively the point drag force is assumed to be weak enough so that the disturbance created by this force fj drag at rj where rj is a position vector to the velocity field at position ri of another object i. So there is a drag force at point j interest is how does it create the perturbation in the velocity at another point i which is in the vicinity of j and that can be approximated by a linear function for small disturbance. So that linear function is given by this vi prime is equal to minus omega ij into fj drag this is omega ij is the second order tensor. So this is a very common framework by which you do. Now of course if you have confinement then some of these basic assumptions may need to be corrected because then when the confinement is very strong you may have to consider some special type of interactions like van der Waals interactions for example. So when you have to consider those kinds of interactions and we will see later on that how those kinds of interactions are taken into consideration. Now one important effect of the confinement is hindered diffusive transport that means confinement hinders diffusion how is it possible presence of the channel wall restricts the movement of large particles thereby affecting their special distributions within the channel viscous interaction between the particle and the wall intensifies for narrower confinements. One may calculate hindered diffusion of Brownian particles by closed form expressions or analytical expressions under certain assumptions. The molecular dimensions much much greater than solvent molecule dimension wall is smooth on the particle length scale length of the channel is large enough to neglect end effects particle concentration is sufficiently dilute to neglect particle-particle interactions a well defined potential field between a single particle at R inside the channel and the wall exists. But you can see some of these assumptions are violated in many practical scenarios like for example particle concentration in many times is highly concentrated it is not dilute. So then it is very difficult to obtain analytical expressions for the effective diffusion coefficient and one has to go for numerical techniques to do that. The next consideration is very important that intermolecular forces that intermolecular forces over a few nanometer length scales tend to become very important and we will discuss about this intermolecular forces. Wicker than the repulsive electrostatic forces the intermolecular forces that may be attractive in nature are responsible for fluid molecules sticking together water wetting glass water not wetting Teflon and a lot of other stuff that we observe every day. Intermolecular forces are weaker than ionic or covalent bonds and intermolecular forces are responsible for the physical state of a compound solid liquid or gas. Some examples of intermolecular forces van der Waals interactions arise from two kinds of forces which can be mirrored back to electrostatic interactions one is dipole interactions another is dispersion interactions and of course you can have hydrogen bonds. So dipole interactions are electrostatic interactions between oppositely charged regions of polar molecules which are called as dipoles. You can see that the water molecule how it can be treated as a dipole I mean this is well known to you from your basic knowledge in chemistry. So whatever we are going to discuss now will be an interface between fundamental physics fundamental chemistry and engineering. Dispersion interactions dispersion forces are caused by the motion of electrons and increase as the number of electrons increases and dispersion forces are the weakest of all intermolecular forces. I mean these may be weak but over certain scales that you are considering these may play their roles van der Waals interactions exist between any two bodies irrespective of their size or polarity this is very important. They are long range interactions where the interaction energy varies as 1 by r to the power 6 with separation distance of r between the interacting bodies. Three types of van der Waals forces occur namely orientation force which is also called as key sum interaction induction force which is called as Debye interaction dispersion force which is called as London interaction. Now I will go through this in a summarized manner. If you are interested to learn more of in great details about all these interactions I can refer to a very good book which is intermolecular and surface forces by israeli. So if you read that book you will be able to get a whole lot of details which I will summarize here. Orientation force or key sum interaction these force results from the electrostatic interaction between two polar molecules. Example water molecules both having freely rotating permanent dipole moments. So the orientation force or key sum interaction concerns electrostatic interaction between two molecules both of which are polar. This is the keyword both are polar. The interaction is obtained by averaging over thermally excited probability distributions of rotation angles. And attractive force is generated when the dipoles adopt anti-parallel arrangement since this is energetically more favorable and repulsive force when they adopt parallel orientation. And the formula for the van der Waals orientation force is given by this. So I will not get into the much details of the formula because we are not going to derive this. And just in case you require to use such a formula the most important key aspect to keep in mind is that it scales with 1 by r to the power 6. Induction force or Debye interaction how is it different from the previous one? This force results from the electrostatic interaction between a polar molecule with a freely rotating permanent dipole moment and a non-polar molecule. So one polar molecule the previous was one polar with another polar. Now one polar with another non-polar averaged over the thermal thermally excited probability distribution of rotation angles. The electric field from the permanent dipole moment induces a dipole moment in the non-polar molecule as mu induced is equal to e into alpha. Alpha being the polarizability of the non-polar molecule okay. So the non the electric field from the permanent dipole moment can induce a dipole moment even in a non-polar molecule and that is how it gets polarized. So it is eventually a polar-polar interaction but the second polar was induced polar it is not the originally polar. These 2 dipoles interact electrostatically to give the induction force which again scales with 1 by r to the power 6. See all these scales with 1 by r to the power 6 because fundamental is dipole-dipole interaction. Originally there are not 2 dipoles but another dipole is induced in this case. The third one is the most interesting which is called as dispersion force or London interaction. This force results from the induced dipole-dipole interaction of 2 non-polar molecules. So the big question is if they are non-polar how do you have dipole-dipole interaction? So the answer is like this. A non-polar atom has an instantaneous dipole moment which can be associated with instantaneous positions of the nucleus around the atom. The electric field from this dipole polarizes any nearby atom inducing in it a dipole moment. The resulting electrostatic interaction between these 2 dipoles correspond to exchange of virtual photons generating an instantaneous dispersion force. So it is an instantaneous phenomenon but effectively this is also dipole-dipole interaction and then it scales with 1 by r to the power 6. Particle wall electrical double layer interaction. So we have already discussed what is electrical double layer. So you can have a charged interfacial layer in vicinity of the wall as well as in vicinity of the particle and let us say there is a particle which has come in close proximity to a wall. So if the particle has come in close proximity to a wall the particle has its own electrical double layer. The wall has its own electrical double layer. Now because of the interaction between these 2 electrical double layers and when they come very close to each other they may be strongly overlapping electrical double layers. So that can give rise to a net force and one can derive complicated expressions for the potential of interaction of the ideal field of 2. One is the particle of 2 entities one is the particle and another is the wall. And this is a function of the zeta potentials of the wall and the particle surface and the Debye length and the separation of the particle from the wall. Solvation interaction. When a flat surface is introduced into a liquid these are all very intermolecular interactions. So very interesting to study these. When a flat surface is introduced in a liquid which is very common like if you have a say solid boundary and if you are putting the solid boundary in contact with a fluid. The liquid molecules adjacent to the surface rearrange to pack well against it as this helps to lower the overall free energy. This ordering of liquid molecules does not require any liquid-liquid or liquid wall interaction. It is solely determined by the geometry of the molecules and how they pack around the constraining boundaries. Similar ordering also occurs for solvent molecules around the particle or solute molecules in a solution. Forces arising from any disruption of such ordering are known as solvation forces. So I will give you a more detailed accounting or origin of solvation interactions. When one solid surface is brought in vicinity of another one. So you have basically let us say that you have a parallel plate channel. So think the situation in this way you had one plate and you had the liquid molecules surrounding that plate. So there was a particular ordering. Now when another plate comes in close proximity of the first plate to form a very narrow channel then the ordering of this liquid molecules is disturbed. Ordering of the liquid molecules around the first plate that is disturbed and they rearrange to find the most energetically favourable packing density. So it takes energy to disrupt the original ordering of the liquid. Thus the presence of the second surface generates an oscillatory solvation pressure. It is whether oscillatory or monotonic it depends on scenarios. This oscillatory solvation pressure is an example d being the separation between the 2 surfaces. In case the interacting surfaces are atomically smooth. Now there are real engineering surfaces which are atomically smooth. It might be surprising but there are such surfaces like mica is an example of an atomically smooth surface that is atomically smooth. So for such surface the solvation interaction can be approximated by an exponentially decaying cosine function. The expression is given where the result is normalized in terms of the molecular length scale. So d by am. In case the surfaces are randomly rough the oscillatory nature is smeared out and it appears to be a monotonic solvation pressure. So only an exponential function without the cosine part. So the cosine part is there if it is a smooth surface. It is not there if it is a rough surface. Structural interactions. Now layering of solvent molecules in the vicinity of a solid substrate means that the liquid layer exhibits a solid like behaviour with a particular lattice constant. This is a remarkable phenomenon. So you we know that solids have a structure and they tend to if they have a crystalline structure they tend to assume a lattice constant. On the other hand liquids do not have such a structure but close to a solid the liquids tend to assume a structure which is influenced by the structure of the solid. So if the lattice constant of the liquid which is layering on the solid is not equal to that of the solid then the bonds between the solvent molecules in the solvation induced layer structure will be strained. This is because of the mismatch of the lattice constant and the interaction energy resulting from some such straining is known as structural interactions. Origin of the layered structure is caused by the exponentially decaying solvation effects. It is inferred that the structural interaction energy density is expressed in this particular form which is based on the form due to solvation interactions. You see this e to the power –d by a m but it depends on the elastic modulus of the solid form of the solvent. For example when the solvent is water the elastic modulus e is that of ice and the value is like 17 giga Newton per meter square. So like these are these are very non-intuitive physical understanding at least to people like us who have our domain in the engineering scenario this is something which is not very very intuitive to deal with but if you are dealing with nanotechnology these are certain aspects that you should keep in mind. Now so with on the basis of several physical issues I mean I have discussed about one or two physical issues but there could be several more. On the basis of these physical issues or not necessarily on the basis of these physical issues but on the basis of the general scenario in the nanofluidic domain we will try to revisit first the Lucas-Warsburn equation okay. So if you recall that for capillary filling dynamics we attempted to give an accounting of the forces acting on the system by using a lumped mass approach. So if you look into this equation you have the inertia that is the mass into acceleration right hand side the resultant force surface tension viscous and gravity. Now if you neglect the gravity and inertia and then if you also neglect the variation in the contact angle if you consider uniform viscosity neither slip nor stick at the interface then you can arrive at this very simple expression that relates that basically equates the viscous force in a fully developed flow. So we assume also fully developed flow in a capillary the fully developed flow with the surface tension force. So a little bit of different notation that we have used here we have used gamma for surface tension coefficient and eta for viscosity. Now the first effect that we may consider in the nanodomain is slip and how does this equation get affected in presence of slip. So let us go to the board and try to see. So assume that there is a channel a circular channel of radius r and let us say that vz is the axial velocity. So now we are interested to find out the viscous drag force when there is a slip at the one assuming fully developed flow. So if it is a Newtonian fluid 1 by r d dr r d v z dr is equal to 1 by mu dp dz. If it is a Newtonian fluid and fully developed flow other terms are not there you have the pressure gradient term and the viscous term. So now if we integrate this equation so r d v z dr is equal to 1 by mu dp dz r square by 2 plus c1. Now we can apply the boundary condition that at r equal to 0 dv z dr equal to 0 that will give you c1 equal to 0. So you have dv z dr is equal to 1 by 2 mu dp dz into r. So vz is equal to 1 by 4 mu dp dz into r square plus c2. Now what is the other boundary condition? What is the boundary condition at small r is equal to capital R? vz is equal to if we write this is it correct? If we write this vz is equal to ls into dv z dr at the wall of course that is of course is it correct? Actually there is an error in it so I want to pinpoint that error. So when you write v is equal to l dv dn or u is equal to l du dn that n is the outward normal. So if you consider this wall if you want to write vz is equal to ls dv z dn that n is this direction and our r is opposite to that so that has to be adjusted with a minus sign. So you can write 1 by 4 mu dp dz r square plus c2 is equal to minus ls by 2 mu dp dz into r. So what is c2? c2 is minus ls r by 2 mu dp dz minus 1 by 4 mu r square dp dz. Now we are interested to calculate the wall shear stress. So what is tau wall? Again because of the same reason we have a minus sign minus mu dv z dr at small r equal to capital R. So that is equal to minus half dp dz into r from this equation. But dp dz is not known. So we will express the dp dz in terms of the average velocity. So how do you get the average velocity?  We have replaced the c2. So what is v average? Integral vz into 2 pi r dr from r is equal to 0 to capital R by pi r square.  So v average is equal to 1 by 2 mu r square dp dz then r cube dr is r to the power 4 by 4 minus r square into r dr is r square by 2. So r to the power 4 by 2 then minus 2 ls r into r dr is r square by 2. So r cube by 2.  So you can cancel this r square. So this becomes r square, this becomes r. So minus 1 by 8 mu dp dz sorry minus 1 by 8 mu dp dz into r square plus 4 ls r. So you can write dp dz is equal to minus 8 mu v average by r square plus 4 ls r. So tau wall is equal to minus half dp dz into r square plus 4 ls r. So you can write dp dz is equal to minus 8 mu v average by r square plus 4 ls r. So tau wall is equal to minus half dp dz into r which is this equation. So 4 mu v average by r plus 4 ls alright. The drag force is tau wall into 2 pi r l.  Typically for practical considerations ls is significantly less than r. So this is like 1 plus x to the power minus 1. So this can be approximated as 1 minus x by binomial expansion if this x is small. So this is as good as approximately 8 pi mu v average l into 1 minus 4 ls by r.  If l is the instantaneous location of the meniscus. So let us get back to the slide to get that picture. So you can see, so whatever we have considered as mu as viscosity in the board the same is there as eta in the slide. So 8 pi eta l dl dt into this correction factor 1 minus 4 ls by r is equal to 2 pi gamma r cos theta okay. So this is the first modification. You can say that this is the very obvious modification. So what is so special about this modification? A very subtle thing is that we do not know what is ls. See it is not a mathematical exercise that you simply assume some value of ls. Typical slip lengths are that may be few nanometers. So you take some 5 nanometer, 10 nanometer whatever and then do this exercise. So this particular formula is of course for small ls by r but even if ls is large you can keep this that in the denominator by keeping instead of 1 minus 4 ls by r in the numerator you have 1 plus 4 ls by r in the denominator that much you can do. But the question is how will you get ls? So to get ls one has to effectively do molecular dynamic simulations. I will not get into the molecular dynamic simulations today but I will show you some results from that which can be used to calculate the slip length. But before doing that I will consider some other approaches that is instead of bringing it through the slip length obtained from molecular dynamic simulations. People have tried several other means to adjust for the observed phenomenon in the molecular domain or in the nanoscale domain. So in the nanoscale domain the Lucas Washburn equation without slip many times fails to capture the essential physics. So instead of taking slip as a alternative measure because slip length is not known people have taken other ways of modifying the Lucas Washburn equation to capture the trends which match with the reality. So one is the inertial so one includes the fluid mass to enter the capillary at t equal to 0 in the LHS. This is the added mass of the virtual mass that we had discussed earlier. And the inertial terms may be included often as an adjustable parameter to explain the dynamics. In the Lucas Washburn equation or the Lucas Washburn model typically we do not have the inertial effects because it is viscous versus surface tension. So inertial term may be used as an adjusting or fitting parameter to match the observed phenomena with the behavior from the equation. So that is one group of researchers have done this. Then dynamic contact angle so we have already discussed that dynamic contact angle may be important question is how important is it in the nanoscale. So typically people have tried to fit the dynamic contact angle as a function of the static contact angle the surface tension coefficient, the viscosity, the slip length and the instantaneous velocity. Or people have used the molecular kinetic theory root to express the dynamic contact angle in terms of the wetting line friction but it is something which is relatively more abstract than the previous approach. But all these approaches have shortcomings we will discuss about that. Then viscous resistance fully developed Poiseuille flow is assumed to obtain the viscous term and modifications are incorporated to include the effect of entrance and the meniscus. The stick or slip associated with the flow also affects the viscous term. So many times this effect has been taken into account by using a modified viscosity. So instead of using the original bulk viscosity the modified viscosity is used as a function of the original viscosity the instantaneous displacement of the meniscus l and l dot. So this has been done in one of the papers and slip length has been used as an adjustable parameter to fit the data but not from fundamental calculations. So this has been reported in one of the earlier papers. Now the what remains as an unaddressed facet of the Lucas-Walburn equation is that if you use these kinds of modifications the modifications in the dynamic contact angle would suggest progressively less value of the contact angle as the meniscus velocity decreases. This we have already seen because like if you recall the dynamic contact angle scales with capillary number to be power one third right. So if you have lower velocity due to viscous resistance you have the contact angle also less as the because theta will scale with capillary number to the power one third. So if theta becomes less cos theta becomes more and the driving force becomes more that means you could see an increasing trend in the l square versus t characteristics. This is what is what is predicted by the standard dynamic contact angle model. But actually we observe this we observe a decreasing trend in l square by l square versus t over the scales that we are considering and not that increasing trend what we see over some elevated length scales. So the question is where is the fallacy and how can we account for that by combining the Lucas-Walburn modified Lucas-Walburn theory with the molecular dynamic simulation results. We will take up this issue in the next lecture. Thank you very


Welcome everyone to the lecture number 8 of non-linear dynamical systems. So, we will see some further extension of the Lyapunov's theorem in different contexts. So, we will begin with global asymptotic stability. So, what is global about the asymptotic stability of an equilibrium point? If an equilibrium point is asymptotically stable, then we know that inside a small neighborhood, inside some neighborhood it is in fact the only equilibrium point. We cannot make such a statement for stable equilibrium points, but for asymptotically stable equilibrium points we know that there is a neighborhood such that all trajectories beginning within that neighborhood all of these converge to that equilibrium point. Hence, that equilibrium point is the only equilibrium point inside a neighborhood. Hence, this is also locally asymptotically stable, but we are curious if this equilibrium point is a globally asymptotically stable equilibrium point. In other words, no matter where the initial condition is, all these trajectories starting from arbitrary initial conditions, all of them do they converge to the same equilibrium point asymptotically? This would make it a globally asymptotically stable equilibrium point. Can Lyapunov's theorem tell us something about this? So, for this purpose we need to see what is the region of attraction. So, suppose this is an equilibrium point and trajectories are coming to this inside some neighborhood. We would like to make this neighborhood larger and larger. In the context of equilibrium in the context of asymptotically stable equilibrium point, we can speak about a region of attraction. No matter where the initial condition starts inside this region, this equilibrium point is attractive. It attracts all these initial conditions so that the trajectories converge to this equilibrium point. So, we would like to look for a larger set that is such that all initial conditions inside this set converge to the equilibrium point. So, we can speak of the region of attraction which is the largest set of all initial conditions such that starting from this set, starting anywhere from this set, trajectories converge to the equilibrium point. So, the region of attraction being the whole space is what makes that equilibrium point globally asymptotically stable equilibrium point. Of course, this rules out any other equilibrium point. So, of course, when there are some more equilibrium points, then the equilibrium point cannot be a globally asymptotically stable equilibrium point. That time the region of attraction is something genuinely difficult to calculate, but Lyapunov's theorem can give us some estimates. So, let us go back to that particular proof of the Lyapunov's theorem. This was our equilibrium point and we had taken this set omega beta. We had taken a value of beta and we had taken all those particular points whose value is less than or equal to beta, where all those points where the Lyapunov function has value at most beta. Of course, we also saw that if we take a larger beta value, then this omega beta could be will be larger because it of course contains this lower value beta set. This is omega beta 2 and earlier one was omega beta 1 because beta 1 is less than beta 2. This omega beta 2 set will contain omega beta 1 set and we also saw that this omega beta set is positively invariant and hence starting inside this trajectories will remain inside this set. Moreover, if the Lyapunov function is strictly decreasing at all points except the equilibrium point, then anywhere it starts inside this omega beta set, it converges. So, we could consider taking larger and larger values of beta and consider the corresponding omega beta set. This particular Greek alphabet is omega, it is capital omega, unlike the smaller omega we have seen. This capital omega is what we also use for ohms. So, if we take a beta 3 value that is very large, then the corresponding omega beta 3 set could look something like this which is itself another estimate of the region of attraction. So, can we take this beta set larger and larger and construct this omega beta sets? That is the next question. So, this cannot be done for all Lyapunov functions unfortunately, but whenever we can find the beta value such that the omega beta set is a bounded set, it is automatically closed by the definition, but it is also bounded set this is not guaranteed for large values of beta. If it is a bounded set, then that particular set is guaranteed to be a region of attraction. So, let us see what can happen for large values of beta? Why would this not be a bounded set for larger values also? So, this particular this is the equilibrium point its interior and this particular contour itself is what we call omega beta set. Here only the contour now a larger value of beta will have a contour that encircles this that encloses this, but for this beta value a little larger it is possible that this contour does not close. It could go and become very large. We will see some problems about how this beta for this contour sets corresponding to larger values may not close. So, this is some possibility that can happen because of which our region of attraction estimate cannot be obtained from the omega beta set. So, this was suppose for beta equal to 1, this one was for beta equal to 4, while for beta equal to 8, for beta equal to 8 suppose it turns out that this particular contour does not close in this direction nor in this direction. So, if it is not a closed contour then that omega beta set would not be a compact set and we will not be able to use our result that this compact set is positively infinite hence the solution exists for all future time. Due to these difficulties we are not able to get a good bound for the region of, good estimate for the region of attraction. So, what do we do in this case? Can we have a Lyapunov function that rules out such contour sets? That is the next question we will ask. So, what is the problem before we formulate the property of such Lyapunov functions? So, notice that when we take, when we go along this direction as we go further and further from this equilibrium point the value of beta is not increasing. For example, if this contour does not close for beta equal to 8 then no matter how far you go along this direction beta equal to 8 value is never reached. Similarly, in this direction beta equal to 8 value is not reached when we go arbitrarily far along this direction here. However, when we go along this direction along this particular radial direction the beta value does increase. It could become arbitrarily large it appears along this direction but not in this direction. This is what motivates us to define a property called radially unbounded. So, as I said level sets of the Lyapunov function give us some estimates of the region of attraction. This could be possibly conservative in the sense that the region of attraction might be larger but this level set gives us only a conservative estimate, gives us a smaller estimate this is possible. So, the next question is can we ensure that the Lyapunov function has some property such that these close sets, these contour sets will automatically close will be bounded. So, this is what we will call radially unbounded. So, we will call a Lyapunov function v radially unbounded if along any radial direction v becomes unbounded. In other words whenever the norm of x becomes very large which means that we are going very far from the origin. If we are going far from the origin then we are going very far from any other finite point also. Whenever we are going very far then it implies that the Lyapunov function value also becomes arbitrarily large. So, this is what we with this property that the norm becoming large implies that the Lyapunov function also becomes large is called radially unbounded. So, it is unbounded along every radial direction. This is precisely the problem we had seen in the previous contour set because of which contours were not closing. However, if we ensure that this Lyapunov function is radially unbounded then it is ruled out that some contour will not close will not be a closed curve. So, not closed contours for any level set is are all ruled out by ensuring that this Lyapunov function is radially unbounded. So, what is the Lyapunov theorem on global asymptotic stability? This is also again a sufficient condition. Let x equal to 0 be an equilibrium point for this differential equation x dot is equal to f of x and let v be a function r into r which is continuously differentiable. Continuously differentiable functions are also called c1. Why? Because they are differentiable ones and the derivative is continuous. Suppose there exists a function v that is continuously differentiable such that v of 0 is equal to 0 and v of x is greater than 0 for all non-zero x. Of course, this is like we saw before v dot of x is strictly less than 0 for all non-equilibrium for all points other than 0 and this important third property that the Lyapunov function v is also radially unbounded. Then the point x equal to 0 is globally asymptotically stable. It is asymptotically stable was already implied by the first two of these three conditions by including the radially unbounded condition. It is also globally asymptotically stable. So, it is not just asymptotically stable but in fact globally asymptotically stable. Let us now come to the case of linear systems. Why? Because this is one situation where the asymptotically stable equilibrium point is in fact globally asymptotically stable. Also, this is a situation where the condition is Lyapunov's theorem is not just sufficient but also necessary. So, a matrix A, a square matrix A is called Hurwitz. If all its eigenvalues have their real part negative. So, if all the real parts, if all the eigenvalues are in the open left half complex plane then that matrix A is called Hurwitz. So, x dot is equal to A of x. A x has the origin as an equilibrium point certainly. This origin is an asymptotically stable equilibrium point if and only if the matrix A is Hurwitz. This is a standard result for linear systems. For linear systems, what happens to the Lyapunov's theorem? We will very soon state a theorem that says that this Lyapunov's theorem is necessary and sufficient. So, what is the result? Consider the system x dot is equal to A x where A is n by n matrix. Then the matrix A is Hurwitz if and only if there exists a function v from R into R which is continuously differentiable such that v is 0 at the equilibrium point and it is positive everywhere else and its rate of change is strictly decreasing. Its rate of change is strictly negative. In other words, Lyapunov function is strictly decreasing along every trajectory. So, please note that we have if and only if here which means that the equilibrium point is asymptotically stable if and only if such a Lyapunov function exists. In fact, the Lyapunov's theorem can be utilized in a more nicer way where the rate of change, rate of decrease can be specified. For linear systems, take v of x equal to x transpose px for some symmetric matrix P. So, a symmetric matrix P is called positive definite if x transpose px is greater than 0 for all non-zero vectors x. So, v of x is a positive definite function if and only if the matrix P is a positive definite matrix. So, what is the rate of change? Rate of change also turns for linear systems where the Lyapunov function v of x is a quadratic function coming from a symmetric matrix. It turns out that the rate of change is also again coming from such a symmetric from another symmetric matrix x transpose qx. This we will verify very quickly. So, it turns out that for linear systems for every prescribed rate of decrease, for every prescribed rate v dot of strict decrease of energy, then in fact exists an energy function v that decreases at this prescribed rate. This is possible such a Lyapunov function v exists that decreases at a prescribed rate if and only if the linear system is asymptotically stable. So, this is something about solvability of a so called Lyapunov equation. So, consider the system x dot is equal to a of x for a square matrix n by n, then a is Hurwitz if and only if for every symmetric matrix q less than 0. So, what was less than 0? We will quickly review the definition of less than 0. Less than 0 is just a negative of greater than 0. So, q being negative definite, negative definite just means that minus q is positive definite. So, a is Hurwitz if and only if for every q that is negative definite that exists a matrix P such that P is symmetric and positive definite. So, please note that we have used the word positive definite only for symmetric matrices that is this is the main convention. If and only if there exists a P such that P is greater than 0 and a transpose P plus P a is equal to q. So, equation 1 says that the Lyapunov function is positive which is which is required for a Lyapunov function and the second condition says that the rate of change of the Lyapunov function is equal to x transpose q s x which is which is guaranteed to be negative for all non-zero x because q was a negative definite matrix. So, this is something we will quickly verify. So, consider the system x dot is equal to a of x in which the Lyapunov function is defined as x transpose a x sorry x it was defined as x transpose P x where P was positive definite where already assume that P was symmetric which means P is equal to P transpose. So, what is the meaning of P is equal to P transpose in the in the context of this function because this is a scalar this is just a real number x transpose P x transpose of this is same as the number itself. What happens when we take transpose of such a product we can take transpose of all the individual elements and also reverse the order.  So, that leaves only x.  Hence, we could as well assume without loss of generality that P was symmetric otherwise we could have considered it is so called symmetric part. So, this is our Lyapunov function we have already assumed P to be symmetric in this case. Let us see what happens when you differentiate the Lyapunov function with respect to time here we get by by the rule when we have a function depending on x on a product of such functions we will differentiate this.  In this particular term we have differentiated this x in this particular term we differentiated this x whether you take the derivative first and then take the transpose or you take the transpose and then differentiate it is the same. So, this d by dt of x is nothing but a times x this is our dynamical system.  So, we see that this can be simplified as x transpose something in the middle and x you can pull out this x this particular term evaluates to x transpose A transpose P x and this is nothing but the x here has gone here and this x transpose has come here and as I said this is nothing but a transpose P. So, the rate of change of this Lyapunov function we have prescribed as equal to x transpose Q x. This is what we want. What is that Q? It is precisely A transpose P plus P A. If P was symmetric you can check that this particular matrix inside the bracket here is also symmetric. You can take transpose of this whole thing and see that you get back the same two terms. So, it is guaranteed to be symmetric and it is precisely equal to V dot of x that is equal to x transpose Q x. So, now in other words prescribed V dot of x equal to x transpose Q x our rate of change of the Lyapunov function is negative for all nonzero x which is nothing but the definition of Q less than 0 Q negative definite matrix. If this is prescribed it means that A transpose P plus P A is prescribed to be equal to Q and of course the dynamical system A is already given. Can we find P such that this equation is satisfied after we are given with A and Q? To be given with A is just to be given the dynamical system while to be given Q is to be given a prescribed rate of decrease of the Lyapunov function. So, given A and Q in this matrix equation each of these three matrices A, P and Q are n by n matrices in which P and Q are symmetric matrices. If A and Q are given can we find a P such that this equation is satisfied that is the first question. This particular equation is called the Lyapunov equation. This is called motivated by Lyapunov's theorem of course Lyapunov equation. It is a matrix equation. So, we are interested in solubility of this equation given A and Q find a P. Moreover, we are also interested can we ensure that this P that we have found also satisfies P positive definite that is what we want from Lyapunov functions. They should also be positive definite functions. So, to this particular result it says that you can find such a P if and only if the matrix A was herwits. So, let us look at this particular equation slide again. So, consider the system x dot is equal to A of x for A and n by n matrix. Then A is herwits which means all its eigenvalues are in the left half open left half plane if and only if for every prescribed v dot in other words for every Q negative definite there exists a matrix P such that P is positive definite and P satisfies the Lyapunov equation. In other words A transpose P plus PA is equal to Q. So, we are now back to non-linear systems. We have seen what happens about Lyapunov theorem for linear system there it was necessary and sufficient. So, we are here to the situation of non-linear systems again. So, here it was only sufficient. So, we take a candidate v of x greater than 0 and we check if along the trajectories of x dot is equal to f of x whether v dot of x is less than 0. In case v dot of x is not less than 0 then there are two possibilities. The equilibrium point is not asymptotically stable. This is one possibility. Other possibility is that we did not choose our candidate v of x carefully which means that we should look for another v. We might have some physical reason some intuition why the differential equation equilibrium point is in fact asymptotically stable, but our candidate Lyapunov function did not satisfy strictly less than 0. Hence, we should not go ahead and conclude that it is not asymptotically stable because less than Lyapunov function not negative definite implies any one of these two possibly both. Whether it can be possibly both or not is a good exercise to think about. So, because the Lyapunov theorem is only a sufficient condition for either stability or asymptotic stability it can be any one of these two cases. So, an important question is what about the converse? If we knew for sure that the system is asymptotically stable then can we say that there should exist a Lyapunov function for the case of linear systems they have already seen. So, under some conditions on f of x and knowing that the equilibrium point is already stable by knowing by some other means whether it is possible that we know that the equilibrium point is stable or asymptotically stable can we guarantee the existence of a Lyapunov function. So, one way to do it is since we already know that we can find Lyapunov functions for linear systems the linear system that is obtained by linearizing the non-linear system at an equilibrium point that linearized system also can be studied for studying the stability of the equilibrium point. So, suppose a suppose small a is an equilibrium point which means that f evaluated at a is equal to the 0 vector define a of define the matrix a as the Jacobian matrix of f at the point a at the point x equal to a. So, we have we will very soon see a big matrix definition of this del f by del x evaluated at x equal to a. So, consider the linear system z dot is equal to az what are the equilibrium points for this z equal to 0 is certainly one equilibrium point it is the only equilibrium point we have seen if and only if a is a non-singular matrix if a is herwits then the linear systems equilibrium point it has only one equilibrium point if a is herwits why because if a is herwits all the eigenvalues are in the open left half plane in particular no eigenvalue can be at the origin and hence it is non-singular then the linear system has only one equilibrium point z equal to 0 and we have already seen that this equilibrium point is asymptotically stable if and only if a is herwits then we can use the conclusion from a we can use a Lyapunov function from a for the non-linear system also. So, let us see this in little more detail. So, f was a map from R n to R n a is a so called Jacobian matrix of f evaluated at a. So, what is the Jacobian matrix the first row is to differentiate the first component of f first with respect to x 1 then x 2 etc. Second row of the matrix a is the derivative of f 2 with respect to x 1 x 2 etc. Similarly, we construct the last row of the matrix a as the derivative of f n because f was a map from R n to R n f itself had n components we call them f 1 up to f n this Jacobian matrix itself is a function of x 1 up to x n we evaluate that matrix at x equal to a then it becomes a constant matrix for this constant matrix we consider its eigenvalues and check whether they are herwits whether this matrix is herwits or not. So, z of the linear system I am sorry about this mistake here. So, we need here z dot is equal to a of z z dot is equal to a z. So, z of the linear system is like deviations from a. So, z equal to x minus a. So, a is herwits implies that the point x equal to a is an asymptotically stable equilibrium point for the non-linear system x dot is equal to f of x. What about the Lyapunov function for the non-linear system Lyapunov function for the linear system already exists we know because for linear system the Lyapunov theorem is necessary and sufficient. So, for the non-linear system also we will consider using this Lyapunov function. So, the Lyapunov function for the linear system whose existence was guaranteed is also a Lyapunov function for the non-linear system. This is a great part about non-linear systems and we linearize if the linearized system is asymptotically stable then the non-linear system also is asymptotically stable and we can in fact use the Lyapunov function from the linear system as Lyapunov function for the non-linear system also. Thus, the linearized system asymptotically stable implies that the non-linear system also is asymptotically stable at that equilibrium point. At another equilibrium point we might consider again linearizing whose eigenvalues may or may not be in the left half plane. As far as this equilibrium point is concerned which when we linearize we get a matrix A that is her weights that equilibrium point is what we will call is so called exponentially stable. This is some good type of asymptotically stable equilibrium point which is automatically satisfied for linear systems. Whenever the linear system is asymptotically stable then that equilibrium point is also exponentially stable. We will see this in little more detail later. So, as I said this Lyapunov theorem is only a sufficient condition. If one particular candidate fails to be less than or equal to 0 or strictly less than 0 then we might consider another Lyapunov candidate. But after some time when various candidates do not work we could consider trying to prove instability using one or more of these candidates. So, here we have a sufficient theorem on for instability. So, what should that function we satisfy so that we conclude instability. After we have failed to conclude stability for some time we could consider instability proving since none of our candidates worked. So, this is again a sufficient condition. So, let x equal to 0 be an equilibrium point for the differential equation x dot is equal to f of x and let v be a map from R n to R. Suppose it is continuously differentiable and v of 0 is equal to 0. Suppose v x naught is greater than 0 for some point x naught that is having arbitrarily small length. So, no matter what length no matter how small a positive length somebody specifies suppose we can find the x naught point whose length is smaller than that particular specification such that v of x naught is greater than 0 at that point x naught.  Suppose it turns out that this function v of x is increasing inside this set u then this x equal to 0 is an unstable equilibrium point. So, if we can show that inside some set u that comes arbitrarily close to the equilibrium point 0 if this Lyapunov function and also the rate of change of the Lyapunov function are both positive then this particular equilibrium point which we initially aimed for proving stable is now unstable. So, this is a sufficient condition to show that a equilibrium point is unstable. So, the fact that v is positive and v dot is positive there is nothing very important about positive both could be negative also because then we could have taken minus v instead of plus v. So, if for some point sufficiently close to an equilibrium point a function v and v dot have the same sign and are both non-zero then that particular equilibrium point is unstable. This is the same theorem reverted in plain words. So, now that completes Lyapunov theorem and its various extensions. There is one important extension called the LaSalle's invariance principle that we will now see. Before we go to this particular principle we will see an example. So, consider the system x dot is equal to this particular example. For example, this example comes when studying the differential equation of a pendulum. Suppose there is this pendulum and there is this deflection theta and this is how gravitation acts here. So, then we have two states let us call this x1 equal to theta, x2 equal to theta dot, x2 equal to theta dot is nothing but sorry this x2 we want to call as theta dot which is nothing but x1 dot. Notice that this equation is precisely the first equation that is written here. So, also this is a pendulum with some friction. In other words whenever it is moving at some velocity theta dot in the time there is some rate of decrease of x2 dot coming from frictional force. This d times x2 has an interpretation that there is some deceleration because of friction. In other words the second differential equation says that this is coming from just one second order differential equation. So, this particular differential equation is what we are going to study now in the context of Lyapunov's theorem and also the next principle that we will see. So, from physical principles we can consider a Lyapunov function as this is nothing but our potential energy plus the kinetic energy. Assume for now then the kinetic energy is just taken as theta dot square. Whether this is a good Lyapunov function or not we will decide soon. Of course, I have written theta and theta dot here. We will write this back in terms of x1 and x2 1 minus cos of x1 plus x2 square. So, let us see what happens to v dot of x. v dot of x is rate of change of this with respect to time. So, this is nothing but d by dt of minus cos of x1 plus d by dt of x2 square. So, derivative of cos of something is minus sin of that. So, this becomes plus sin x1 but x1 itself is a function of time and we were differentiating with respect to time this times x1 dot plus 2 times x2 times x2 dot and this is the rate derivative of this term. So, now we will use x1 dot and x2 dot from our dynamical system equations and put that here.  This was x2 dot. So, when we evaluate this it seems like sin x1 times x2 is also here but with a factor 2 and it is not cancelling well. So, we are not able to conclude. So, this whole thing is equal to minus x2 sin x1 minus 2 bx2 square. Of course, the frictional element b is such that b is greater than 0. So, this term is helping us to show that this less than or equal to 0. Unfortunately, this one is this guaranteed to be always less than 0? Can we say that x2 times sin x1 is always positive? Can we say that x2 times sin of x1 is always greater than 0? If we could say this, then this would always be less than 0 and we would show that the pendulum is as stable as x1, x2 equal to 0, 0 as a stable asymptotically stable equilibrium point. In fact, we are not able to say this. Of course, that only means that this, that could mean that this Lyapunov function was not a good candidate. But then we could also go back and see that look, this was not really kinetic energy. While this is potential energy, this is not kinetic energy, we could consider dividing by 2. So, let me make a change on the same slide with a different color pen. So, here we have this divided by 2. We have this again divided by 2. This cancels off this 2. This cancels off this 2 also, because of which we perfectly have a cancellation here now, because of which this 2 is gone, but that is not a problem. So, now we have that this is equal to, we will see the same example in a different context also. But at least here it appears that v of x equal to 1 minus cos of x1 plus x2 square, not a Lyapunov function. But v of x equal to 1 minus cos of x1 plus x2 square by 2 has v dot of x equal to minus bx2 square, which is less than or equal to 0. Hence, equilibrium point 0, 0 is stable. Of course, we should have checked at the equilibrium point, whether 0, 0 is an equilibrium point to begin with. One can verify in the system of equations that 0, 0 is indeed an equilibrium point. That is why we are trying to study whether this equilibrium point is stable, unstable, asymptotically stable. So, the pendulum with some friction has been shown to be a stable equilibrium point. Is this asymptotically stable? Can we say that this is strictly less than 0, whenever x1 and x2 are both not equal to the 0, 0 point. Can we say strictly less than 0 for the second candidate? And the first one was not a Lyapunov function for the second one. In other words, if we say v dot of x equal to 0, does that imply that when would we say that this Lyapunov function is, rate of change of the Lyapunov function is negative definite, if the only point where it was equal to 0, where the rate of change was equal to 0, happened at 0, 0. So, if b x2 square minus of that is equal to 0, this is a question mark here. It only implies that x2 is equal to 0. We cannot say anything about x1. So, x1 can be arbitrary. In other words, when the pendulum is not moving, when the pendulum is stationary, at all those places, the Lyapunov function rate of change is equal to 0. The Lyapunov function is not strictly decreasing at those points, but that does not imply that the x1 component itself is also 0. Only the x2 component is 0. So, there are various points where the rate of change of the Lyapunov function becomes equal to 0. It is not necessarily only the equilibrium point. And hence, this Lyapunov candidate does not help us to prove asymptotic stability. However, do we think that the equilibrium point is asymptotically stable? We could consider linearizing it at the equilibrium point and checking. So, equilibrium point is x1, x2 equal to 0, 0. So, we can differentiate this and construct our matrix A. The first function is not a function of x1. Hence, we have 0 here. This is a function of x2. Rate of change of this, partial derivative of this function with respect to x2 is equal to 1. Second function, this is our f2. So, this is equal to f1 of x1, x2 and the second one is equal to f2 of x1, x2. So, we are going to use our big matrix that we saw in the slide here. The component, the value that comes here is derivative of f2, partial derivative of f2 with respect to x1. In other words, derivative of this with respect to x1. So, this is equal to minus cos of x1. And what comes here is derivative of this f2 with respect to x2, which is equal to minus b. And this is expected to be a function of x1 and x2. So, we have to be evaluating at x1, x2 equal to 0, 0. So, when evaluated at this point, we get this equal to 0, 1, minus 1, minus b. Cos of 0 is equal to 0. Cos of 0 is equal to 1. So, what about this matrix? Does it have eigenvalues in the open left half plane? At least our Lyapunov function could not prove it is asymptotically stable. But the linearized system, let us investigate the eigenvalues of the linearized system, eigenvalues of A. In other words, eigenvalues of this matrix assuming b itself, the friction is caused by some b that is positive. Only then it is deceleration due to friction. So, eigenvalues of A can be found by finding the determinant of S i minus A. In other words, determinant of S minus 1 plus 1 S plus b. This is equal to S times S plus b plus 1. This is S square plus b S plus 1. So, what can you say about the roots of this polynomial? Roots of this polynomial is precisely the eigenvalues of this matrix A. We have constructed the characteristic polynomial and found the determinant of the S i minus A matrix and its roots, roots of S square plus b S plus 1. Of course, we could use the quadratic equation and find the roots. By using that b is positive, it will turn out that the roots are in the left half complex plane. Why? The roots have product equal to plus 1. We could evaluate this so that we do not need to discuss in detail why the roots are in the left half plane. Roots are equal to minus b plus minus square root of b square minus 4 Se. So, that is precisely b square minus A and C are both equal to 1 divided by 2. So, suppose roots need not be real. Roots could be either real or complex. If the roots are complex, which means b square is less than 4, then because b itself is positive, then roots in we will call this open left half complex plane. If the roots are complex, complex roots, then would we have complex roots if this discriminant is negative? In other words, b square is less than 4. That is when this quantity within the square root back under the square root sign becomes negative. In that case, the imaginary part is positive for one, negative for the other. But the real part is minus b by 2 and we already assumed that b is positive and hence clearly the roots are in the left half complex plane and they are complex. But what about when the roots are real? That means b square is greater than or equal to 4. When b square is greater than or equal to 4, then can we say that square root of b square minus 4 in absolute value is less than or is strictly less than b, strictly less than b, again absolute value. Why? Because you see this b square is greater than 4 and this is some number from which we subtracted 4 and then taking the square root and it cannot be even equal to b, it will be strictly less than b. In other words, this minus b plus minus square root of b square minus 4, this quantity itself in absolute value will be less than 0. When b square minus 4, we take positive root and we add that to minus b without the absolute sign. So, minus b plus b square minus 4 is also negative. Why? Because b square minus 4 quantity itself is less than b in absolute value. Of course, when both are negative, it will only be further negative. So, whether as long as b is positive, whether the roots are real or complex, they are in the left half complex plane. So, again, roots in c minus. So, we have investigated both the cases, whether the roots are real or complex. We know that the eigenvalues of that matrix are in the open left half plane. Open meaning, it is not on the imaginary axis also. It is strictly in the left half complex plane. So, what did we conclude from this? Linearized system. A linearized system for this differential equation has origin, that is, which is asymptotically stable. Even though our Lyapunov function constructed from physical principles, what physical principle? We use a Lyapunov function as a notion of energy in which we added the potential energy and the kinetic energy. In the kinetic energy term, we ensured that the number 2 is important. To divide by 2, divide the velocity square by 2 was required. In spite of constructing the energy function from physical principles, we could not prove that the origin is an asymptotically stable equilibrium point. We could only show that it is stable. But the linearized system has eigenvalues that are in the open left half complex plane, because of which the origin for the linearized system is asymptotically stable. So, there should exist a Lyapunov function. This particular symbol just says there exists. As soon as the linear system is asymptotically stable, that matrix A was Hurwitz. Hence, we already saw that there should exist a Lyapunov function, because for linear systems, the Lyapunov theorem is not just sufficient but also necessary. So, if we know that the origin is an asymptotically stable equilibrium point, then there is a guarantee that there is a Lyapunov function. And this Lyapunov function, we can in fact construct as X transpose Px. So, X transpose Px also is a Lyapunov function for the non-linear system. So, we know that there should exist some Lyapunov function, even though that could not be motivated from physical energy principles. There should exist a Lyapunov function which can in fact prove asymptotic stability. So, how do we actually find this? This is something that we can do as an exercise, because we know that for linear systems, we can prescribe the rate of decrease and still be able to find the energy function that has precisely that decrease. This energy function is also guaranteed to be positive definite simply because the matrix A was herwits. So, we will see an example of such a prescribed rate of decrease and solve the Lyapunov equation in an exercise. So, the next thing to do is we could consider using that same Lyapunov function which could prove only stability but not asymptotic stability to arrive at the conclusion of asymptotic stability by using a so called Lasals invariance principle. So, Lasals invariance principle helps us, for example, in this situation where the Lyapunov theorem could only prove stability but not asymptotic stability. It can prove at elsewhere places also. We will see now. So, for example, there could be a situation where we want convergence to not an equilibrium point but to a set. So, what are equilibrium sets? What is the trajectories are converging not to a point but to a set? We could speak about stability in stability for sets also. So far, we spoke only for equilibrium points but we could also consider speaking for sets. So, in this context, there is a Lasals invariance principle. So, let omega be a compact set that is positively invariant. Positively invariant means it is positively invariant with respect to the dynamics of the function f. We are considering the system x dot is equal to f of x here and in that context, we have constructed a set omega that is a compact set. Suppose, you could find a function v from this domain D to R and suppose this function v is C1. C1 we already saw means it means continuously differentiable and this function v satisfies the property that its rate of decrease, its rate of change is less than or equal to 0. So, it is strictly decreasing is not what we are assuming, non-increasing on omega. Now, considering the set E of all those points in this set omega where v dot of x is in fact equal to 0. Inside this set E, we will now look for an invariant set. We will now look for the largest invariant set in E. Suppose, we construct the set M which is the largest invariant set sitting inside E, then Lasals invariance principle says that every solution that starts in omega approaches M as T tends to infinity. As time tends to infinity, it approaches this set M. So, C1 as I said is a function that is continuously differentiable. One stands for at least once. So, what does it mean for a solution to approach a set M? This is something we have not seen yet. So, every solution starting in omega approaches set M as T tends to infinity. What is the meaning of this statement? So, we will see this quickly before we end today's lecture. So, for every initial condition x0 in omega, x of T tends to M. Approaches M as T tends to infinity is nothing but x of T tends to M as T tends to infinity. So, this x of T tends to M means converging to a set. So, for that purpose, we are yet to define what is the meaning of converging to a set M. So, distance of a point P, we can speak of the distance of a point P, not just from another point, but from a set. So, it is the shortest distance of P, shortest with respect to various points Q you can take in that set M. So, consider the set M and distance, this is nothing but distance. Distance of the point P from the set M is the infimum. For infimum, you can think of minimum for the time being. There is an infimum of this norm. You take different, different points Q and find the distance of P, distance between P and Q and you look at the minimum such value as you vary this point Q across the set M. So, the point Q in M which is closest to P we take and we take the distance between P and Q and this distance we will call as the distance of P from the set M. So, to converge to a set M means the distance of x of t of the vector x of t at any time t, distance of that vector from the set M, this distance is some real number now, it is a positive number. It goes to 0 as t tends to infinity. So, you can check that if this point P is sitting inside the set M, this distance is 0. Why? Because it is at distance 0 from itself which is already inside the set. So, it is with respect to this notion of distance of a point P from a set M and this distance going to 0. This is the notion of approaching, a trajectory approaching a set M. Using this, we look back at Lassalle's invariance principle. Suppose these conditions are satisfied that you could find a compact set that is positively invariant with respect to the dynamics of F. If you can find a function V that is continuously differentiable and non-increasing, its rate of change is less than or equal to 0 on this set omega. Inside this set omega, you construct the set of all points E that are such that V dot is equal to 0. Inside this E, you construct the largest invariant set M. Lassalle's invariance principle says that every solution starting in omega approaches M as t tends to infinity. So, we will see that this invariance principle allows us to use that same Lyapunov function motivated by physical principles to conclude in fact asymptotic stability. This we will see in the following lecture. Thank you.