 interfacial waves on time dependent base states. Here the container bottom was being oscillated up and down with an amplitude a and a frequency capital omega. We had analyzed this problem in the oscillating frame of reference where we had found that the effective value of gravity would be g plus a omega square into cos omega t. This led us to equations for the perturbation velocity potential and boundary conditions and we found that the essential difference was that that in one of the boundary conditions we had a time dependent coefficient. The coefficient was time periodic. Consequently when we wrote down our forms for phi and eta, we kept the time dependence arbitrary. We did not set them equal to e to the power i omega t, i small omega t as we have done until now. This is because the equations have a time periodic coefficient or a time dependent coefficient and so e to the power i omega t is not going to work. You can substitute and see that. So we have phi of t and e of t and we had said that we are going to determine what function is phi of time and what function is e of time. Now we have started with substituting these forms into the kinematic boundary condition. This had led us to two equations a and b. Again notice that the important difference is that that these equations, these homogeneous algebraic equations have time dependent coefficients. Similarly the Bernoulli equation gave us an equation coefficient of cos kx has to be said equal to 0 here and the coefficient of sin kx also has to be said equal to 0. This will give me two more equations in C1, G, C2 and H. So we have a total of four equations in four unknowns. So by setting the coefficient of cos kx and sin kx to be 0, we obtain d phi by dt 1 plus exponential minus 2kh into C1 plus G is equal to 0. Let me call this equation C and you can write down a similar equation for C2 and H. The coefficients remain exactly the same plus exactly the same thing. I will call this equation d. So you can notice a pattern here that either I can treat this as a 4 by 4 matrix multiplying C1, C2, G and H or I can take this as a 2 by 2 matrix. The same pattern exists earlier also although we have not explicitly utilized it. You can see that either I can take equations a and c and equations a and c are just in the unknowns G and C1. Or I could take equation b and d and they would be in the unknown H and C2. I am just going to use these two equations and work on them and you will find that if I had used instead this I would have been led to exactly the same conclusions. Let us use equations a and c. So the equations for a and c let me write it down properly. So we have de by dt into g minus phi of t into k into 1 minus e to the power minus 2kh into C1 is equal to 0. This is equation a. I am just rewriting it and then we have equation c. Equation c I am just going to rewrite it so that g comes first. So capital G comes first. So I am going to exchange the two terms into g. So there is a g here first and a g here next and then plus d phi by dt into 1 plus e to the power minus 2kh into C1 is equal to 0. So now you see we are going to analyze these two equations. These are two homogeneous equations in two unknowns g and c1. The only important thing to note is that their coefficients are actually unknown functions of time. This is unlike what we have done until now where in all the problems that we have done until now the coefficients have been constants and the determinant has led us to a dispersion relation. Here we will find that we will have to establish a certain relation between e and phi capital E of time and phi of time and this will render the coefficients of these equations to be independent of time. In particular note that if I set d e by dt to be equal to phi then this equation the coefficients of this equation become independent of time. You can see that very easily that there is a d e by dt here and a phi here. If I set them to be equal then I can write them as either phi of t into g minus phi of t into the rest and then pull the phi of t out that phi of t can be taken out as common phi of t is in general not 0. So it is just the rest of the equation whose coefficients are time independent. So if I make this substitution then the coefficient of the first equation becomes time independent. We will remember this and we will use this. Let us work on the matrix for this it will be a time dependent matrix and we will work out its the usual criteria that the determinant has to be equal to 0. This will ensure this will give us some some equation where there will be both d e by dt and phi and then we will plug this assumption in into that equation. Let us work out the matrix first. So our determinant will have d e by dt then minus phi of t k 1 minus e to the power minus 2 k h and then here there will be g plus a omega square cos omega t to e of t and then the next term is d phi by dt into 1 plus power minus twice k h is equal to 0. This gives us the equation d e by dt into d phi by dt to 1 plus exponential minus 2 k h plus g plus a omega square cos omega t into phi 0. This is our equation. Clearly you can see that there are two unknowns here e and phi both of which are unknown functions of time. As I argued earlier if I set phi is equal to d e by dt. I have said this earlier that we are going to set phi is equal to d e by dt and this first of all it renders the first equation its coefficients become independent of time. So, if I set phi is equal to d e by dt then you can see immediately that you have a d e by dt here. This becomes d square phi by dt square into 1 plus e to the power minus 2 k h and there is a phi here and phi is d e by dt. So, I can pull out a d e by dt and keep it as common and then the phi he has already gone out as d e by dt and then what is left I am just writing is equal to 0. In general d e by dt is not equal to 0 e of t is not equal to 0 d e by dt is not equal to 0 at all times. So, this is not equal to 0. So, the only other option to satisfy this equation is to put the quantity in square brackets to be 0. But this has the effect that it gives us an equation governing e as a function of time. Let us see what kind of an equation are we getting. So, I am just setting the coefficient I am setting the quantity inside the square bracket equal to 0 and this gives me d square e by dt square. There was a coefficient 1 plus e to the power minus 2 k h I am dividing both sides by this quantity. So, this quantity is no longer there. So, this quantity will shift to the second term and the second term will become g k plus a k omega square cos omega t. There was a 1 minus e to the power minus 2 k h here and I have divided out by 1 plus e to the power minus 2 k h. So, this put it in a bracket and then this is e of t is equal to 0. I can simplify this further and write this in a slightly more compact form d square e by dt square plus you can see that this is nothing but if I multiply up and down by e to the power plus k h then this is e to the power plus k h minus e to the power minus k h divided by e to the power k h minus plus e to the power minus k h and this is nothing but tan hyperbolic k h. This is the familiar tan hyperbolic k h which has appeared before when we studied waves where there was no oscillation of the bottom. Now, there is a oscillation of the bottom but the tan hyperbolic is still appearing. So, I am going to write the tan hyperbolic. So, this is this is basically tan hyperbolic. So, I am just going to write tan hyperbolic before the curly braces and then just put it inside. I can write this in a even more compact form once I recognize that g k into tan hyperbolic k h is just my dispersion relation for free oscillations when there is no vibration of the container bottom. So, if I call this omega square then what I have pulled out is g k because g k tan hyperbolic k h we have seen earlier is the dispersion relation for surface gravity waves when there is no oscillatory motion of the bottom. So, I will call this omega square and I have pulled out a g k. So, then there is one here and if I pull out a g k then what I am left here is a omega square by g that is a non-dimensional number into cos omega t into e of t is equal to 0 where I have said omega square is equal to g k this is defined as tan hyperbolic k h. You can immediately recognize what equation is this. This is our familiar Matthew equation. We had studied this when we had looked at oscillations of the pendulum. Now, we are taking our container containing a liquid and a free surface in the base state and we are shaking it up and down and we are asking that if we put a perturbation eta on the surface what is the equation that governs the amplitude of the perturbation. Notice that all whatever forms we have chosen for phi and eta are in the standing waveform the space and the time part are separate. So, this is the equation that governs the coefficient e of t for eta and this is the Matthew equation. We have looked at this equation in some amount of detail before and now we are again finding it. You can now go back and check what is the implication that e is governed by this equation on these two homogeneous equations whose coefficients gave us the equation governing e. The choice d e by d t is equal to phi already made the first equation have coefficients which are not dependent on time. You will see that this second differential equation that we have got that basically ensures that the second equation here equations becomes the same as the first equation which is equation a. So, in other words these two choices d e by d t is equal to phi and the fact that this e itself is governed by a Matthew equation makes both of these equations time independent equations both equations a and c the coefficients it makes it time independent. So, we are essentially finding that the amplitude of the interface which is governed by e of t is actually governed by a Matthew equation. Let us see some limits of this and let us first convince ourselves that this is familiar and this reduces to the correct thing that we have already derived. So, first you can see that in the absence of forcing in the absence. So, without without oscillatory forcing this equation just becomes d square e by d t square plus omega square into e is equal to 0. This is just telling us that e is a sine or a cosine function of time with a frequency which is given by the dispersion relation. This is exactly what we would have concluded if we had done a normal mode analysis on the problem setting the oscillatory part to 0. We have done that before and it is recovering our older results. Note also in particular that setting equation a you can immediately see that setting d phi by d t or d e by d t is equal to phi actually makes this equation a as I have said before it makes it independent of time it makes the coefficients independent of time. Let us write the resultant equations this will give us a relation between g and c 1. So, I have d e by d t I am writing phi as d e by d t. So, now both the terms contain d e by d t g minus k into 1 minus e to the power minus twice k h into c 1 is equal to 0 d e by d t is in general not 0. And so my equation a just becomes a homogeneous equation for g and c 1 with coefficients which do not depend on time this is consistent with what I said earlier. This equation also gives me a relation between g and c 1. So, I can write c 1 is equal to 1 by k 1 minus e to the power minus twice k h into g. Similarly, you can use we have used equations a and c I encourage you to go and do the same for equations b and d and you will not find anything you will find exactly the same thing that you have to set phi is equal to d e by d t. And once you said that you will be led to the same method equation as we have obtained here. So, either you can write it as a 4 by 4 matrix or you can write it as a 2 by 2 matrix we could have done the same in the unforced problems also earlier. So, now if you use equations b, so if you use equation b, so this is from a we have not used b. But if you did that, then you would find that c 2 is equal to the same thing you would have found that phi is equal to phi has to be said to be d e by d t and then c 2 is equal to the same coefficient into h. Let us now use these things recall that the form for phi was phi of t into c 1 cos k x plus c 2 sin k x into e to the power k z plus e to the power minus k z minus twice k h. This was the form that we had written earlier. Now we have determined c 1 and c 2 in terms of g and h and we know that phi is equal to d e by d t. So, I am going to replace this as d e by d t and then I will have I am going to replace express c 1 in terms of g using this and c 2 in terms of h using that. So, I will have 1 by k into 1 minus e to the power minus 2 k h that is a common factor and then I will have g times cos k x plus h times sin k x into the same thing this part. So, I am just going to write this part here. Let us simplify this a little bit more. So, phi is equal to d e by d t and I want to multiply and divide numerator and denominator by g. So, I will have a g k here into 1 minus e to the power minus 2 k h and then I have my usual space part which is g cos k x plus h sin k x into e to the power k z plus e to the power minus k z minus twice k h. I am doing this in order to write this problem in a manner which is as close as possible to the unforced problem that we had earlier seen. So, that you clearly see what is the connection between the forced problem and the unforced problem. So, I will write this as d e by d t this is all this is happening in the denominator. So, this 1 minus e to the power minus 2 k h remains untouched I multiply by e to the power plus k h into e to the power minus k h because you have divided by this I have to multiply by this and the rest of the things remain the same not doing anything to them. Now, I will just multiply the this becomes g this becomes g k I will just multiply this term I will just multiply this term 1 minus e to the power minus 2 k h with e to the power plus k h. If I do that then it becomes e to the power plus k h minus e to the power minus k h. I am trying to get a tan hyperbolic k h which is why I am doing this because g k times tan hyperbolic k h is my omega square. So, I have got this I have multiplied e to the power plus k h. So, I have to put a e to the power minus k h e to the power plus e to the power minus k h and then the rest of the part remains the same. So, this just becomes de by dt the e to the power k h here can be shifted to the top and then there is a g and then I can write this as omega square. You can see that this part is not tan hyperbolic h the entire thing inside the bracket is just tan hyperbolic k h. So, that is g k times tan hyperbolic k h we know is the dispersion relation for free oscillations I have written that as omega square. So, and I have shifted the e to the power minus k h to the numerator. So, what am I left with I am left with the remaining which is g cos k x plus h sin k x into e to the power k z plus e to the power minus k z minus twice k h and I have left out this factor here. So, I am going to put this here now you can notice that I can take this and insert it into this I can multiply both the terms by e to the power k h and you will see that the resultant form of the numerator if I divide numerator and denominator by by 2. So, the numerator becomes cos hyperbolic of k times z plus h and the denominator becomes cos hyperbolic of just k h. I encourage you to try this on your own it is a very simple step take this e to the power k h multiply the numerator and then divide the numerator and denominator by a factor of 2. So, multiply the numerator and denominator by half you will get cos hyperbolic k z plus h. So, this resultant expression for phi can be written as phi is equal to d e by d t into g by omega square. Now, there is no exponential because I have shifted the exponential to the cos hyperbolic part the x part remains the same into the x part now the numerator now becomes cos hyperbolic of k into z plus h divided by cos hyperbolic. For reference cos hyperbolic of x is defined as e to the power x plus e to the power minus x by 2. Similarly, eta does not change eta is just we are writing everything in terms of e of t and that is just g cos k x plus h sin. And we have seen that where e of t satisfies d square e by d t square plus omega square 1 plus a omega square by g cos omega t 0. This is our solution to the problem, this is our solution to the problem. What is left behind g h and the initial value of e has to be determined from initial conditions. The main thing that we find now is that unlike earlier is that that now the coefficient e of t is not governed by a simple harmonic oscillator equation. It would have been governed by simple harmonic oscillator equation if we set small a to 0 here. Then this just becomes a simple harmonic oscillator equation whose frequency is just the dispersion relation. The dispersion relation also I will write omega square is equal to g k tan hyperbolic k h. So, we are finding that phi and eta have very similar expressions we have derived these very similar looking expressions we have derived. And we have found so this factor of cos hyperbolic k z plus h divided by cos hyperbolic k h this we have seen earlier in the case of unforced waves. We are seeing the same factor and I have written it deliberately so that it is as close as possible to the previous problem. The only difference that we are finding now is that earlier there was just a dispersion relation. Now, there is an actual equation governing the coefficient e of t. There is only one coefficient now e of t the coefficient the time dependent part of phi is just d e by d t. So, if we know e as a function of t then we also know d e by d t as a function of t. So, this is the Matthew equation and we have looked at its stability chart in one of the previous videos early on in this course. In the next video I am going to analyze this Matthew equation a little bit more and explain what it means for the fluid problem. What does instability mean for the fluid problem and we will also discuss some practical applications of this. It turns out that there are quite a few practical applications of this force vibration problem particularly for atomization applications.