 In the previous lecture, we derived the Boltzmann distribution. So, let us recapitulate a little bit that what was it. So, we expressed the number density of ions. If you look at the final formula in this slide, number density of the ions in terms of the bulk and number density and the potential within the electrical double layer. Now this is a very commonly used form of the ionic number density distribution, but with its simplicity, there are certain assumptions that need to be carefully taken into consideration. So what are the assumptions? In the Boltzmann distribution, ions are considered to be point charges. So, when ions are considered to be point charges, there is no theoretical upper limit of the number of charges which can occupy a surface. However, because of the finite size of the ionic species, you will definitely have restricted number density of ions on a particular surface. The system is in equilibrium with no macroscopic advection diffusion of ions because if you recall that we like we basically recovered the Boltzmann distribution from the North Planck equation by setting the advection terms to 0 and unsteady terms to 0. The solid surface is microscopically homogeneous. So, if you have inhomogeneity then there will be dependence on x also. So, we have considered that the electrical double layer phenomenon is dependent solely on y and not on x. The charge surface is in contact with an infinitely large liquid medium. That means you can apply the first stream boundary condition that at y tends to infinity you have psi tends to 0 that you can consider. Next point very important that when we consider the Boltzmann distribution, we consider the electrical double layer field that is the electric field within the electrical double layer because of the electrical double layer phenomenon. However, you may also apply an external field like let me give you an example by drawing a schematic in the figure in the board. So, let us say that you have a channel like this. Now within the electrical double layer you have a potential field which is psi. Now let us say that to actuate the fluid flow you apply an electrical potential gradient electrical potential gradient along the axial direction. Let us say that is equal to phi not. That potential is phi not. It is a field. So, it is not constant. Basically phi not is a function of x and psi is a function of y. So, question is when we consider the potential distribution and number density distribution within the electrical double layer, we considered only psi but not phi not. But phi not may be there. Even if it is there it is typically not what we have considered. So, the question is why we will answer that now that why we are considering only the electric field within the electrical double layer and not the electrical field that is externally applied. So, consider as an example a typical electrical double layer thickness of 10 nanometer. I will later on show that what are the parameters on which the thickness of the electrical double layer will depend. But let us say that we are considering a typical electrical double layer thickness of 10 nanometer and a surface potential of 100 millivolt, okay. So what is the typical field strength? So, from the surface at the surface the potential is 100 millivolt and in the outside the electrical double layer it is 0 and the distance over which it occurs is 10 nanometer. So, the potential gradient is 100 millivolt-0 divided by 10 nanometer. That much voltage per unit length. So, that is of the order of 10 to the power 7 volt per meter, okay. On the other hand typical axial electric field that is applied for electro osmosis is like of the order of 10 to the power 4 volt per meter. So, the field within the electrical double layer due to electrical double layer is much more over weighing as compared to the field that is externally applied in the axial direction. So, that means when we are writing the Boltzmann distribution ideally we should have considered all the potentials not just the induced potential but also the applied potential. But for practical reasons because the applied potential is much much less than the induced potential within the electrical double layer that is why we are considering only the electrical double layer potential for the Boltzmann distribution. Now the final point which I have already mentioned that the first stream boundary condition is applicable. The first stream boundary condition is applicable means that the electrical double layer I mean is such that as y tends to infinity you have psi tends to 0. So, y tends to infinity mean it is a first stream boundary condition. So, typically you expect that if the electrical double layer is thin then the channel centre line will be a first stream or a located at infinity. So, for example if the electrical double layer thickness is 10 nanometer then 1 micron channel 1 micron height channel the centre line will be at like virtually at infinity. However, the electrical double layer thickness depends on concentration and the channel size also depends on your requirement. So, there may be possibilities when the channel size and the order of the electrical double layer thickness are equivalent. So, then you can have overlapping of the electrical double layers which is a very interesting phenomena but like as a part of this basic course I do not want to complicate matters further by including the electrical double layer overlap phenomena. But electrical double layer overlap may be possible when the channel height the characteristic channel height is of the order of the electrical double of the typical electrical double layer thickness or channel height is less than the order of the electrical double layer thickness typical. But we are not considering such situations. So, what will be the modification in that situation simple modification you at the channel centre line you cannot consider psi equal to 0 you have to consider psi equal to some psi c at the channel centre line which will come from other constraints that what would be psi c. Now, we have to understand one important thing that in the Boltzmann distribution we have related the number density of ions with the potential that is what we have done. But that does not solve the number density or the potential because one is related to the other but you need another equation involving these 2 to close the system and that another equation is fundamentally given by the Gauss law. So, from now onwards we will be referring to a little bit of electrostatics and like I would like to emphasize as I already told you quite a few times that microfluidics is an interdisciplinary subject. So, perhaps many of you like me are not electrical scientists or electrical engineers but neither many of the many of us are fundamentally physicists although we work on areas interfacing physics and engineering. But so it may be possible that many of you are not comfortable with many of the issues that we will be discussing but you know to work on an interdisciplinary subject it is important that we pick up at least the essential matters of interest related to this. And many of these things I will start with the some of the basics of high school physics so that we can relate this with our earlier studies in physics. So, Gauss law so Gauss law in an integral form is like this. So, this is a contour integral of the electric field over ds is equal to integral of rho e total dv the rho e total is the volumetric charge density the total volumetric charge density. So, the surface integral of charge I will tell you what is the physical interpretation of the Gauss law later on. So, I mean when I have separate slides for physical interpretations of all these laws and I will explain more carefully but first let us understand it by simple mathematical terms. So, basically what you are saying that this contour integral is equal to this volume integral. So, then you know you can use this you can convert this area integral or this contour integral into volume integral by using the divergence theorem. So, e dot ds you can write that that is equal to divergence of e right by using the divergence theorem. What is epsilon not epsilon not is the permittivity of the free space right. Now, so because the choice of the control volume is arbitrary that means you can write epsilon not divergence of e is equal to rho e total this is pretty simple and clear. So, both of these are actually equivalent forms this is the integral form of the Gauss law and this is the differential form of the Gauss law and one can be recovered from the other by using the divergence theorem. As we have seen in many cases using the Reynolds transport theorem how can we transform from the integral form to the differential form and all. Now, the total charge density the total charge density is the sum of the free charge density and the bound charge density. So, sum of the charge may be free and sum of the charge may be bound to the surface. So, the bound charge density is defined as minus of divergence of a vector which is called as the polarization density vector. So, the bound charge density is minus of divergence of p where p is the polarization density or the dipole moment per unit volume that originates due to gradients in local dipole densities within the medium. So, now I mean let me try to work out the little bit of algebra that is involved with this in the board. So, we have epsilon not del not e is equal to rho e when we write rho e I am not writing rho e free, but when I am writing rho e please keep in mind that this is free charge density and plus rho e bound, rho e bound is related to the dipole density. So, that is equal to minus divergence of the polarization vector. Now this polarization vector is constitutively related to the electric field through this parameter chi. It is a constitutive relationship between polarization and the electric field because like one is the sort of cause and another is the effect. So, you can write epsilon not del dot e is equal to rho e minus del dot epsilon not chi e. So, you can write del dot epsilon not being a constant you can bring it in or out of the nabla operator. So, it is epsilon not into 1 plus chi this 1 plus chi is called as epsilon r or the relative permittivity and together this is called as epsilon permittivity of the medium. So, eventually you get del dot epsilon e is equal to rho e. So, this is a form of the Gauss law and you can express e as a function of what? Negative of the gradient of the potential. So, you get del dot epsilon minus grad of the potential is equal to rho e. So, that means if epsilon was a constant if epsilon is was a constant in many cases epsilon is a constant then that comes out of the derivative. So, you will get del square of the potential is equal to something in the right hand side. In mathematics if you have del square of something is equal to something in the right hand side that is called as Poisson equation. So, this is also known as Poisson equation if the right hand side becomes 0 it is called as Laplace equation. So, Laplace equation and Poisson equation are very famous forms of partial differential equations in applied mathematics and engineering. So, now we will go to the slide and try to see let us go to the slide. So, you can see that this is the final form that we talked about and what are the boundary conditions we are talking about continuity of the tangential electric field across the interface and jump in the normal flux. So, this is very important see when we talk about electrical field in the tangential direction there may be there should be a continuity in the tangential electric field, but in terms of the normal electric field there is a normal there may be a jump why because across the interface there may be a storage of charge. This is the very basic difference between like say subjects like heat transfer with a subject of say electrostatics. In heat transfer you have heat flux continuous across an interface why because the interface cannot store any thermal energy. Thermal energy can be stored only by a volume, but in electrostatics a surface can store charge because a surface can store charge there can be a normal gradient a jump in the normal gradient of the electrical tangential and that has to be accommodated in the boundary conditions. So, now we have got another equation the Poisson equation and we will combine the Poisson equation with the Boltzmann distribution to get an equation which explicitly which is explicitly expressed as a function of the potential potential in the electrical double layer because one of our objective is to solve for the potential within the electrical double layer. So, the Poisson equation let us start with that. So, Poisson equation del dot epsilon e is equal to rho e. So, what is rho e? Rho e is the total volumetric charge density. So, summation of z i e n i z is the valency e is the protonic charge and n i is the number density of the i h p c. So, we will address a simplified case for this let me come to the board and work it out for you. So, you have del dot epsilon e is equal to rho e e is equal to minus grad psi. So, you have del dot epsilon del psi is equal to minus rho e. What is rho e? Summation of z i e n i. Let us assume as an example, let us assume that there are 2 types of ions. One is z plus positive ion, another is z minus. Let me give you an example. Let us say you have a salt water solution. Typically in electro osmosis many times we use a salt water solution say NaCl water or KCl water. So, after the dissociation reaction, you will have Na plus Cl minus H3O plus OH minus. So, out of that if you have Na plus and Cl minus they are dominating. So, let us say z plus corresponds to Na plus and z minus corresponds to Cl minus. So, then if z plus is equal to z then z minus is equal to minus z for a symmetric electrolyte. This is not always true. It is true for electrolytes like NaCl, KCl where the positive and the negative ions are of symmetric valency. So, for NaCl or KCl, z is equal to 1. So, this is for this kind of electrolytes are called z is to z symmetric electrolyte. Remember the algebra that we are doing is still valid even if it is not a symmetric electrolyte. This is just a example that I am talking about. So, then rho E becomes what? ZEN plus minus ZEN minus. This summation there will be 2 terms plus. So, Ze N plus minus N minus. N plus what is the Boltzmann distribution? N plus is equal to N0 e to the power minus Ze psi by KBT. What is N minus? If this is N plus what is N minus? Yes, that is a common sense answer. Just this minus z will be replaced by plus z. So, N0 e to the power Ze psi by KBT. So, this is N0 Ze. Now, if you divide it by 2 and take the minus sign out, remember e to the power x minus e to the power sorry not divide by 2 multiply by 2 and then divide by 2. Then for the remaining term you know e to the power x minus e to the power minus x by 2 is equal to sin hyperbolic x. So, this will become sin h Ze psi by KBT. So, we have been able to express the right hand side solely in terms of psi and the left hand side is also in terms of psi. So, you have a governing equation for psi. This equation has been obtained by combining the Poisson equation with the Boltzmann distribution. Hence, this is called as Poisson Boltzmann equation. So, the Poisson Boltzmann equation. So, there was a minus rho e that minus and this minus will make it plus. So, 2 N0 sin h Ze psi by KBT. This is the Poisson Boltzmann equation. Now, we will analyze this equation before we solve. We will definitely solve this equation for some special cases to illustrate that what is happening. But before solving, we will try to develop some insight. 1 epsilon I have missed, epsilon is there. Now, for a one-dimensional case, see many times the electrical double layer phenomena gives rise to a one-dimensional potential distribution. Like for example, here if the perpendicular direction of the plate is y, then there is a potential distribution along y. So, then this becomes basically ddy of epsilon d psi dy for 1d. But there are problems where you basically encounter 2d or 3d situations. Take the one-dimensional form and make an order of magnitude analysis of this equation. So, we will make an order of magnitude analysis. Let us say that epsilon is a constant. You may have variable epsilon in a thin interfacial region because of polarization or several other issues. But let us not go into those matters at this stage. Let us assume that epsilon is a constant. So, if epsilon is a constant, then the order of magnitude of the left hand side is what? Epsilon, what is the order of magnitude of psi? See, we are talking about the variation of psi from the shear plane to the bulk. At the shear plane, it is the zeta potential. So, psi zeta divided by what is the characteristic length scale of the ideal? That is the dy length lambda. So, E zeta by lambda square. And the right hand side, the order is 2n0 ze and sinh z psi by kbt. So, z e zeta by kbt. So, that means lambda square is of the order of what? Or 1 by lambda square is of the order of 2n0 z square e square by epsilon kbt. This order of magnitude estimation is very important because this will tell us that how think the electrical double layer will be in terms of an order of magnitude of the characteristic dimension. So, you see, let us consider various parameters in this equation. You will see that in these equations, for a given set of ions, z is fixed. For the fluid, epsilon is fixed. Typically, we are doing these experiments at standard temperature. So, this is fixed and the Boltzmann constant anyway is a universal constant. So, that is fixed. Charge of a proton that is also fixed. So, we can only play for all practical reasons. We can primarily play with n0. So, you can see that 1 by lambda square is proportional to n0. That means lambda is proportional to or lambda is proportional to 1 by n0. So, n by square root of n0. So, what it means is that higher the concentration thinner is the electrical double layer. And typical orders of magnitude, typical numbers, let us go to the slide and see what are the typical values. So, if you see, this is a chart. This chart is prepared by using this formula. So, concentration versus Debye length by taking the standard temperatures and so on. So, if the concentration is 10 to the power minus 6 molar, the capital M as you know is moles per liter. So, 10 to the power minus 6 m, that will give, that is a very dilute concentration. That will give the Debye length around 308.3 nanometer. So, roughly of the order of say around say 500 nanometer. Then 10 to the power minus 5 is like close to 100 nanometer. 10 to the power minus 4 close to 30 nanometer. In this way, 10 to 10 to the power minus 1, you see that how small it is. So, the trick of playing with the electrical double layer thickness is essentially related to varying the concentration of the solution that you are using. So, and the final take home message is that more dilute the concentration, thicker is the electrical double layer. Now, we try to see that what is the solution to this equation. So, the solution to this equation maybe because it is already written in the board, we will refer to the board to do that and then refer to the slide to summarize. So, look at this equation, D Debye or epsilon D2 psi dy2 assuming epsilon to be constant is equal to 2 N0 Ze sin h Ze psi by KBT. So, this is the solution to this equation. This is the equation that we intend to solve. This equation is a nonlinear equation. This is called as nonlinear Poisson-Bordman equation. Then what is the source of nonlinearity? The source of nonlinearity is the presence of the sin hyperbolic term. However, we can start with a simple case when we can linearize this equation, we can linearize this equation when Ze psi by KBT is small. So, for small Ze psi by KBT, small means typically much much less than 1. So, let us say at least 0.1 or less. For small Ze psi by KBT, you have sin h Ze psi by KBT is equal to Ze psi by KBT. Approximately equal to not equal to. For small x sin hx is approximately equal to x. So, small Ze psi by KBT, how will you know that small Ze psi by KBT is small? What is the maximum of Ze psi by KBT? Maximum psi is zeta. So, maximum of Ze psi by KBT is Ze zeta by KBT. So, if z is zeta by KBT is say less than 0.1, we can see, we can approximately assume that it is small. Question is what is the typical zeta corresponding to that? Because we have to relate that with our practical values. So, if we consider z is zeta by KBT less than 0.1 typically, then that zeta will come typically less than equal to or less than, typically less than 25 millivolt. So, if your zeta potential in magnitude is less than 25 millivolt, typically you can apply this linearization. This is known as Debye Huckel linearization. Debye Huckel linearization. So, then you can write this as 2n0 z square e square psi by KBT. So, d2 psi dy2 is equal to n0 z square e square by epsilon KBT psi. What is this? This is 1 by lambda square, right? So, d2 psi dy2 minus 1 by lambda square psi equal to 0. So, this is a very straightforward ordinary differential equation and it has a solution of the form of exponential, right? e to the power x e to the power minus x combination of that. So, that you can write in terms of sin h and cos h also and the boundary conditions are that at y equal to 0 and let us say this is the situation. This is y equal to 0 and this is y equal to 2h. At y equal to 0 psi is equal to zeta and at y equal to 2h also psi equal to zeta, okay. Now, technically there is an incorrectness in giving this boundary condition because what is y equal to 0? y equal to 0 is the, let us say it is the solid boundary. If this is the solid boundary, then zeta potential is not potential at the solid boundary. It is the potential at the shear plane but because the shear plane and the solid boundary is located just at the few angstrom apart and not only that when you are thinking about fluid flow, you are basically giving the boundary condition at the shear plane. So, keeping these 2 factors into account, we approximate this by the zeta potential. So, that is a standard practical interpretation to giving the boundary condition. So, once you give this boundary condition, let us go to the slide to show that like, so you can see this is the equation and boundary condition psi equal to zeta at y equal to 0 and y equal to 2h. So, this is the final solution. You can clearly see that this solution satisfies the boundary condition at y equal to 0 and at y equal to 2h psi equal to zeta, okay. This is one form of boundary condition when you are specifying the surface potential but sometimes it is very difficult to specify the surface potential instead you specify the surface charge density. So, what will happen for that case? We will look into the next slide. Surface charge density is specified at both walls just as an example. So, this how will you relate surface charge with the potential? So, understand the physics. The electroneutrality condition says that surface charge on a wall must be equal and opposite to the total unbalanced charge in the electrical double layer near the wall, right. Because the total in totality it has to be electrically neutral. So, surface charge is minus of the charge in the electrical double layer, okay. So, sigma which is the surface charge density is equal to the minus of the total charge density within the electrical double layer that is the volumetric charge density rho e into dy from y equal to 0 to y equal to h, okay. This is for one plate. There are two plates, right. One plate is influencing between 0 to h. Another plate is influencing from h to 2h. So, for each plate we can write this. So, now where from you can write minus rho e is equal to epsilon d2 psi dy2 from where you can write this? This is the Poisson equation, right. So, in place of minus rho e you write epsilon d2 psi dy2 into dy. So, this becomes epsilon d psi dy. So, epsilon d psi dy at y equal to h minus epsilon d psi dy at y equal to 0. What is epsilon d psi dy at y equal to h? That is 0. Why? Because it is a center line. It is a center line of the channel because both walls have symmetric boundary conditions. So, at the center line you have d psi dy equal to 0. So, you have sigma at y equal to 0. That is the bottom plate is equal to minus epsilon d psi dy at y equal to 0. See, this is the beauty of mathematics and physics. This subject is totally uncorrelated. This is electrostatics. This is totally uncorrelated with fluid mechanics and heat transfer. Look at the analogy of the boundary condition. tau is equal to mu du dy and q is equal to minus k dt dy, the Fourier's law of heat conduction and the Newton's law of viscosity. So, the constitutive behaviors are so similar that you can apply the same mathematical paradigm for applying the boundary conditions. So, just like heat flux q is equal to minus k dt dy that is governed by the Fourier's law of heat conduction. The surface charge density sigma is equal to epsilon d psi dy. So, only thing is that plus or minus sign depends on the coordinate axis. So, for the bottom wall y is directed outward normal to the wall. So, minus epsilon d psi dy for the top wall it is plus epsilon d psi dy because for the top wall the y is acting inward normal to the wall. Now, so if you know sigma at y equal to 0 is equal to minus epsilon d psi dy at y equal to 0 then from the previous expression you can find out sigma and then you can replace basically the zeta with sigma lambda by epsilon. This comes out of the algebra. It is not that it is done ad hoc. So, what you do is in the previous expression whatever was the expression for psi you find out d psi dy and set d psi dy at y equal to 0 and equate that with sigma. So, you will get the zeta potential is equal to. So, instead of psi equal to zeta into cos h d you will get sigma lambda by epsilon into cos h d. That means zeta is replaced by sigma lambda by epsilon and you see there is an intuitive physics behind this. Let us come to the board and explain this. So, you are basically having zeta is equal to sigma lambda by epsilon. So, this comes from algebra but we will bring it from intuitive physics. So, what is the boundary condition? Sigma is equal to minus epsilon d psi dy at y equal to 0. So, let us try to make an order of magnitude. Sigma this is sigma surface sigma s this is sigma s is equal to is of the order of this is epsilon this is zeta by lambda lambda is the d y length. So, you can see that zeta is of the order of sigma s lambda by epsilon and in this case the order of magnitude exactly matches with the exact value. So, this expression is quite intuitive. Now, so we have discussed about potential profiles with d by Huckel linearization for two cases. In one case the potential is specified in another case the charge density is specified. But what about the general case when the linearization is not possible? When the linearization is not possible in general you cannot get the analytical solution except for some special circumstances. So, we will discuss about that special circumstances here. So, maybe to explain this it will be useful for you if we at least work out one or two basic steps in the board and then we will go to the slides. So, we will start with the non-linear version of the Poisson-Boltzmann equation make an attempt to solve it. Many interesting features will come out of this solution as we will see. So, first we will non-dimensionalize the equation. So, to non-dimensionalize let us say y bar is equal to y by H. So, we non-dimensionalize with the half height of the channel. You could as well non-dimensionalize with lambda there is no problem. This is just another non-dimensionalization I am showing. This is also I have important non-dimensionalizing parameter because it happens to be the characteristic system length scale. And what matters is how this system length scale compares with the d y length. That is what is governing the physics of the problem. So, and psi bar let us say this is the. So, first of all this must be a non-dimensional number. Always whenever there is e to the power x, x must be non-dimensional. Because see you expand the e to the power x in series power series. So, different powers of x they correspond to the same dimension only when each is dimensionless. So, whenever you write e to the power x until and unless you write empirical equations never write e to the power x with x dimensional. So, e to the power x means x must be non-dimensional. So, this is non-dimensional therefore this you can interpret as psi divided by some psi reference. What is that psi reference? k B T by z D. This is called as thermal voltage. So, for algebraic simplicity we will see that it will help us in making some manipulations instead of z D psi by k B T we will make z D psi by 4 k B T. This is just for algebra nothing more than that. So, we can write epsilon 4 k B T by z E h square sin h. What is this? 1 by lambda square. So, you have an expression where automatically see the physics governing the problem h by lambda comes into the picture. So, h by lambda we call say k. So, this is d 2 psi d y 2 is equal to k square by 4 sin h 4 psi. Now, d 2 psi d y 2 we can write these are all non-dimensional d d y of d psi d y right. So, this you can write d d psi of d psi d y into d psi d y. You can write d of d psi d y into d psi d y is equal to k square by 4 sin h 4 psi d psi right. So, this is like V d V. If you call this as V then this is V d V that means V square by 2 right. So, that means what you get is d psi d y square by 2 is equal to k square by 4 sin h 4 psi by bar will become cos h 4 psi bar by 4 plus some constant c 1. So, you can basically multiply both sides by 2. So, this becomes 8 plus the nu nu c 1 ok. So, we will. So, this of course you can take a square root after giving a suitable boundary condition and then you can proceed, but to do that you must know what is c 1. To know what is c 1 basically what you have to specify? You have to specify a location where you know both psi and d psi d y right. This is psi bar. So, we assume this is psi that at the center line both psi and d psi d y are equal to 0 right. At the center line both psi and d psi d y are equal to 0. If that is true then you can get c 1 is equal to minus k square by 8, but the question is we have come to a very interesting situation in mathematics. Let us try to understand this carefully. Originally what are the boundary conditions? Let us say you have the channel this is the center line. So, at the wall what boundary condition you are giving? Let us say zeta potential not exactly at the wall, but at the shear plane, but let us say it is the y equal to 0. y center line which is y equal to h if it is symmetric you have d psi d y equal to 0 true. This is a second order ordinary differential equation right the Poisson-Boltzmann equation. So, it requires two boundary conditions. So, these two boundary conditions should have been sufficient to solve the problem, but here you are imposing a third boundary condition which is psi equal to 0. So, that makes it a over post problem. So, a second order differential equation you are using three boundary conditions. So, where is the fallacy? Actually this is not an additional boundary condition this is called as gauge boundary condition. So, what is this? So, basically all the potentials are relative. So, there is a center line potential and relative to the center line potential you are finding actually all other potentials. So, you can set the reference potential as the center line potential with the reference equal to 0 that is all. So, that does not mean that you are using it as an additional independent boundary condition. It is just like if you use p equal to p atmosphere to calculate the gauge pressure similarly it is a gauge boundary condition based on which you calculate the potential distribution. So, but there are situations when this gauge boundary condition cannot be imposed like for example, you do not know what is the psi center line and it depends on other parameters of the problem. So, then you can say that like this kind of treatment will be will not be appropriate, but assuming that the gauge boundary condition is valid we will go to the slides now and see that we will continue with this c 1 equal to minus k square by 8 that c 1 is the constant of integration then these psi dy square is equal to this. Now, when you take the square root you have to consider that whether you have to take the positive root or the negative root and for that just use a physical judgment that if zeta is negative then psi will increase from a negative value at the wall to 0 at the bulk right. So, positive d psi dy must be there. So, that for that you actually extract the negative sign. So, that d psi dy becomes positive if psi is negative. Then what you do is that you use an identity this is just simple mathematical manipulation there is no physics behind this that use this hyperbolic identity that this you can write as sec h square psi by tan h psi okay. This is just like an identity which you can use. So, and you know I mean out of these 2 one you can write that the tan h the derivative of that is sec h square. So, you can write this as log of tan h psi and therefore tan h psi will become exponential of minus k y. Then you use the third so called third boundary condition that at y equal to 0 psi equal to zeta. So, then this is the form that you get that tan h psi is equal to tan h zeta into e to the power minus k y. So, psi is equal to tan h inverse tan h zeta into e to the power minus k y. This is the analytical solution to the non-linear Poisson-Boltzmann equation with certain simplifications. Similar with the just a change of coordinate for the top wall. So, this is valid for y non-dimensional y between 0 to 1. This is for the bottom plate and for the top plate this is the potential distribution. So, we are in we are assuming that the bottom plate is influencing the electrical double layer phenomena adjacent to it and the top plate is influencing the electric electrical double layer phenomena adjacent to that. So, that we are assuming that there is no electrical double layer overlap then both plates will influence both the electrical double layer phenomena and that is one of the assumptions behind the derivation of this equation. Although research has shown that this equation may be valid for weakly overlap Tdl that is the electrical double layer overlap is not strong. There is slightly slight overlap of the electrical double layer and this equation is still valid. Anyway, we will stop our discussion now and we will continue with the electrokinetics in the next lecture. Thank you very much.