 We were looking at the derivation of the Stokes wave. We had seen that first we obtained the equations at linear order these turned out to be homogeneous equations the right hand side was 0. Then we went to order epsilon square the next order and then we found that at this order the equations became inhomogeneous this is similar to what we did earlier in the course for non-linear oscillators except that now we are dealing with partial differential equations. Recall that we have the Laplace equation at every order we also have two versions of the Bernoulli equation the second version of the Bernoulli equation is obtained by taking the total derivative on the Bernoulli equation and then using the kinematic boundary condition to obtain another equation. So, we had two versions of the Bernoulli equation the Bernoulli equation itself and a modified Bernoulli equation using this we went up to second order where we found that the equation governing phi 2 the second order velocity potential phi 2 was governed by again a Laplace equation and then there were two other boundary conditions but now these were all applied at z is equal to 0 still but they were inhomogeneous and what appeared on the right hand side were all quantities which came from the previous order. So, quantities like phi 1, eta 1 and their various derivatives. So, this is what we have found so far. So, let us proceed further. So, let me summarize the equations that we have found until now. So, at order epsilon we have found the equations are our familiar equations at z is equal to 0 I am going to skip writing the z is equal to 0 because at every time we are going to write apply it at z is equal to 0 only. So, I am just going to indicate that by a 0 and then a modified Bernoulli equation this whole thing also at 0 is equal to 0 and we had called this a 1 b 1 and c 1. a 1 is my governing equation b 1 c 1 are really boundary conditions. At order epsilon square we had your square phi 2 equal to 0 then we had del phi 2 by del tau also at 0 plus eta 2 is equal to minus del square phi 1 by del z del tau at 0 into eta 1 and then we have a modified Bernoulli equation the whole thing applied at z is equal to 0 all of these are applied at z is equal to 0 recall that the z is equal to 0 condition is applied only to phi and its derivatives eta whether it is eta 1 eta 2 eta 3 and so on these are not functions of z. So, it is only on phi 1 phi 2 phi 3 and their various derivatives that the z is equal to 0 condition applies also at 0 and we had we can call this a 2 b 2 and c 2. We had understood in the last video how did we get the right hand sides it is by repeated Taylor series expansion and making sure that we have everything up to order epsilon square. As I mentioned earlier a 1 b 1 c 1 are homogeneous whereas in order epsilon square b 2 and c 2 are inhomogeneous. Note that on the right hand side of b 2 and c 2 what appears are all terms coming from the previous order. So, just like whatever we have done up till now using perturbative techniques we have to solve at any given order before we can proceed to the next order. So, here also we will have to solve the equations at order epsilon and then use them in order to simplify the equations at order epsilon square the solution that will appear at order epsilon will be used to determine the right hand sides of the equations at order epsilon square and the same pattern will repeat. Again note that the left hand side the operators which appear on the left hand side of the equations both at order epsilon and order epsilon square are very identical. It is just the right hand side which keeps changing and in this particular problem we will have to also write down the equations at order epsilon cube. We will do that shortly but note that the right hand side will just keep getting more and more lengthy. We will keep adding more and more terms on the right hand side whereas the structure of the left hand side the left hand side operator will remain identical at every order. Also note that the quantity omega 2 has not yet appeared in our equations. Recall that omega 2 was I expect omega 2 to give me my frequency correction. How does the frequency depend on amplitude and so omega 2 should appear in a non-linear at a non-linear order. However, order epsilon square on the right hand side of the equations you can see that there is no omega 2 as yet. So, in order to determine omega 2 we will actually have to go to order epsilon cube. We will do that shortly. Let us now solve the equations at order epsilon first. Let us write down the solutions and use it to simplify the equations at order epsilon square. So, order epsilon so this is solution. The solution at order epsilon is easy. We are basically looking at traveling wave solutions. This entire procedure can also be done for standing waves but let us work on traveling waves. So, at order epsilon our solution is familiar. So, in non-dimensional way it is e to the power z sin of x minus and eta 1 is cos of x minus tau. This is our solution to the equation. We are not writing down the most general solution to the equation but you can go and check that these satisfy the equations that we have written earlier. Recall that we have written the equations at previous order are grad square phi 1 equal to at this order are grad square phi 1 equal to 0 del phi 1 by del tau at 0 plus eta 1 is equal to 0 and del square phi 1 by del tau square plus del phi 1 equal to 0. You can go back and check that these two satisfy these equations. This is this should be clear where we have got these solutions from. If you go back into one of the early videos in the course where we discussed traveling wave solutions you will find that we have written solutions of this form. Again whether eta 1 is cosine or whether phi 1 is sin that depends on which term we had written it we had written the most general solution and you can go back and see that one specific instance of the most general solution is what we have written here. And you can see by substituting and checking that the solution that we have written is a solution to the equations that a1, b1 and c1. So, these are the equations a1, b1 and c1. So, we have found the solutions to equations a1, b1 and c1 that we already know we are just writing it in a non-dimensional sense. So, that is why instead of writing e to the power kz we are writing e to the power z, z is a non-dimensional distance. Similarly, kx is written as x and then there is a omega t which is also non-dimensionalized. So, now this tells us the solution at this order. So, now let us go to the next order. So, order epsilon square. So, at order epsilon square we of course have the Laplace equation which is grad square phi 2 is equal to 0 that is homogeneous. So, there is nothing to be done there. But the other two boundary conditions have to be simplified because what appears on the right-hand side of those boundary conditions equation b2 and c2. So, equations b2 and c2 you can see that the right-hand side depends on phi 1 and its various derivatives as well as eta 1. We now know what is phi 1 in eta 1 and so we need to substitute and work out the precise functional form of the right-hand side of equations b2 and c2. Only then can we solve these equations. So, our next task is to work out the functional forms of the right-hand side. Let me give some numbers to these terms. So, I will call this term 1 and this is term 2. So, now I am looking at term 1 and term 1 is minus del square phi 1 by del z del tau at 0 into eta 1. This is just what I have written as term 1 in green. So, I need to work out the form for that term and this you can show easily from knowing these two solutions. This you can show easily is just cos square x minus tau. Similarly, term 2 is minus half del phi 1 by del x whole square plus del phi 1 by del z whole square evaluated at 0, z is equal to 0 and term 2 will just turn out to be minus half. There will be a cos square x minus tau and a sin square x minus tau they will add up to give you 1 and so you will just be left with the minus half prefactor. So, the right-hand side side of B2 equation B2 is cos square x minus tau minus half which I can write it as. So, cos square if I multiply it and divide by 2, then I can write it as 1 plus cos 2x minus 2 tau 1 plus cos 2 theta is 2 cos square theta is this and so this just becomes half cos 2 of x minus. Notice the appearance of this quantity cos 2x minus tau, notice the appearance of this quantity. We had started with a cos x minus tau here and at non-linear order on the right-hand side what has appeared is a cos 2 times x minus tau. Why 2? Because there was a product between two things here. This is coming from the quadratic nonlinearity and that is causing a cos square to appear here and which in turn we have written it as. So, cos square theta we have written it as cos 2 theta. So, we have a cos 2x minus tau appearing in our equations and you can see that that is basically a non-linear effect. So, if you put a travelling wave of the form cos x minus tau then what appears due to nonlinearity is cos 2 times x minus 2 times tau. We will discuss this more. Let us now proceed and similarly you can write down the equation C2. Similarly, we need to write down the right-hand side of equation C2 and so equation C2 is basically the modified Bernoulli equation 0 and this if you work it out. So, once again you can label them. So, I will label them here. So, I will call them term 3, term 4 and term 5. So, you will find the term 3 and term 4 once you substitute and work out the forms of term 3 and term 4 you will find that 3 and 4 will cancel each other whereas 5 will just be 0. So, overall the right-hand side of equation C2 is just 0. So, please work this out, try this yourself. So, 3 will be equal to 4 and 5 will be equal to 0. 5 is 0 because there is a del phi 1 by del x whole square plus del phi 1 by del z whole square evaluated at z is equal to 0. That will eventually add up to a sin square plus a cos square and that will give you 1 and del by delta of a constant is just 0. So, you will get 5 equal to 0 and consequently the right-hand side of equation C2 will be 0. It will just become a homogeneous equation. So, equation C2 is just 0. So, now we have this thing where at order epsilon square we have grad square phi 2 is equal to 0. That is all Laplace equation. Then we have del phi 2 by del tau at 0 plus eta 2 is equal to half cos twice x minus tau. This we have already found before and then we have del square phi 2 by del tau at 0 plus del phi 2 by del z at 0 is equal to 0. The 0 is a consequence of cancellation. So, these are all my updated equations a2, b2 and C2 where we have worked out the right-hand sides for b2 and C2. And you can see that it is only for equation b2 that there is a right-hand side. Equation C2 does not have a right-hand side, it is just 0. So, you can see that the equation governing phi 2 is this and that. So, you can think of it as this is a boundary condition, this is an equation governing phi 2. And if you know phi 2, you can use equation b2 to determine eta 2. We are going to choose because the right-hand side of equation C2, this equation is 0. We are going to choose phi 2 to be equal to 0. You can think a little bit more as to why this choice is justified. If we make that choice, then equation this obviously satisfies the Laplace equation. So, we do not have a problem with that. And now we just need to determine eta 2. Eta 2 is just determined from this equation. Phi 2 is identically 0. So, all derivatives are 0. So, eta 2 just becomes half of cos 2x minus. So, I will put that also in a bracket. So, you see what has happened is this is telling us that at non-linear order, we have started at linear order with a linear traveling wave solution, something that we have already seen previously in the course. And we know what is the velocity potential corresponding to a linear traveling wave. We also know what is the dispersion relation. What we are finding at quadratic order is or at non-linear order is basically that there is a second harmonic which appears at the interface eta 2. Whereas, there is no correction to the velocity potential at this order. Let us see what happens. So, now you see this you can think of it as phi is equal to epsilon phi 1 plus epsilon square phi 2. This is what we had written. Then we have eta is equal to epsilon eta 1 plus epsilon square eta 2. And we also have a tau which is t into 1 plus epsilon square omega 2 plus dot dot dot. And there are other here also. So, you see we have, we know this, we know this, we also have determined this, we also have determined this. But we do not yet know the value of this. So, our problem at order epsilon square or problem at order epsilon square is still incomplete. We need to determine this. As I had mentioned earlier, omega 2 has not yet appeared in the equations at order epsilon square. Recall the equations A2, B2, C2. On the right hand side, we did not have a small omega 2 appearing anywhere in those equations in the equations B2 and C2. So, we now need to proceed to the next order and do the same exercise at the next order and see whether omega 2 appears at the next order. We will find that omega 2 does appear at the next order. We are not going to solve the equations at that order. We will just find that there are some resonant forcing terms which appear at that order. We will have to get or eliminate those terms, the resonant forcing terms just as we had done that for a nonlinear pendulum or a nonlinear oscillator. Pay attention that now we are doing this in the context of partial differential equations. Earlier when we are looking at nonlinear oscillators with single degree of freedom, whether it was duffing oscillator, whether it was a nonlinear pendulum, we were looking at ordinary differential equations. Of course, in the method of multiple scales, we had converted it into equivalently into partial differential equations, but these are really partial differential equations because now we are looking at a fluid system. We are looking at what is the nonlinear modification to a linear traveling wave. We are trying to calculate that and we have calculated what is the nonlinear modification. A harmonic of the primary wave that we have imposed is appearing, the second harmonic. But we have not yet determined the correction to the frequency of the wave or in other words, the correction to the phase speed of the wave which is determined by this small omega 2. For this, we need to proceed to the next order and see how to determine omega 2 from that order. So, let us continue. So, at order epsilon cube, once again we will have the Laplace equation. Now, for phi 3, our Bernoulli equation will pick up a number of terms on the right hand side. I am just going to write those terms and then I will explain a few of them as to how it works. The idea is similar to what we had done earlier, but now we have to go to a few more terms in the Taylor series expansion. So, I will pick up one term and that will lead to three terms which appear on the right hand side. I will explain how those three terms have come and very similar manner, you can understand how the remaining terms have come. There are a number of terms on the right hand side of this equation. So, let me write those terms first. So, minus, so I will put a bracket because all the terms on the right hand side are minus. So, minus eta 2 del square, everything again gets applied at 0. This is just a consequence of the Taylor series approximation. The right hand side now becomes quite lengthy. So, it will take some time to write it. So, those are the terms on the right hand side. You can see that this is quite lengthy. There are a total of 8 terms which have appeared. Now, before we start working on them, we have to do the same thing as before. We have to work out the forms of each of those terms from what we know about the previous order. In particular, there is a simplifying feature at the previous order, phi 2 is 0. So, some of these terms will go to 0. So, I am just going to set those terms to 0 where phi 2 appears. So, we have this term which is 0, we have this term which is 0 and we have this term which is 0. So, that eliminates 3 terms from my right hand side and I am left with total of 5 terms. Now, I will just explain, I am just going to put in boxes some of these terms. So, because there are so many terms, so it is a good idea to explain where they have come from and how do we write this Taylor series expansion that omega 2 something that we wanted. This is the reason why we came to this order and this has appeared now at this order. This was this had not yet appeared at order epsilon square. Now, we have a small omega 2 on the right hand side. So, let us see. So, recall that in the Bernoulli equation, we have a term in the Bernoulli equation before we did all the expansions, we had a term del phi by del t at z is equal to eta. We express del by del t in terms of del by del tau, del by del t was this into del by del tau and phi itself was expanded as epsilon phi 1 plus epsilon square phi 2 and this whole thing was applied at z is equal to eta. So, I am going to show you how this expansion can be used to generate those terms, this, this and that. So, let us see how. So, we will have 1 plus epsilon square omega 2 plus let me not write the dot because that is not important here and then we have epsilon del phi 1 by del tau. Now, this will be applied at z is equal to eta, but now I am going to do a Taylor series expansion on this. So, del phi by del tau at z is equal to eta is del phi by del tau at z is equal to 0 plus because there was a multiplying factor. So, I again put an epsilon here, the next term would be the second derivative of the derivative with respect to z of this quantity again applied at z is equal to 0. So, it will become a second derivative del z into del tau once again at 0 and now I have to multiply it with eta, eta itself is epsilon eta 1 plus epsilon square eta 2. This is just coming from Taylor series expansion of this term Taylor series and I have gone up to the second term in the Taylor series expansion. Note that all the derivatives are evaluated at z is equal to 0 even when we go up to non-linear order. So, this is the first term. Similarly, we can do a similar exercise with the second term. So, let us do that and so we will have, so now let us work on the, so I have we have missed out one more term. So, we will need to write the Taylor series approximation up to one more term. So, this is the next term would be half and then there would be a del cube phi 1 by del z square into del tau. So, this is just the del by del z of this into eta square and eta is basically epsilon eta 1 plus epsilon square eta 2. I do not want to go up to a order epsilon to the power 4. So, I am just writing eta as epsilon eta 1 and there is a factorial 2 here which I have put here and this is whole squared. So, this is whole squared. So, all of this, this, this and this is just coming from this term alone and I will put a bracket here.