 In the last problem, we had looked at solving for roots of a quadratic equation, where the equation was slightly perturbed from a state where it was factorizable as two integers, two integer roots into a slightly perturbed state. There I had stated that the epsilon which is typically a non-dimensional parameter and a small parameter for solving the problem perturbatively is introduced through non-dimensionalization. We did not discuss non-dimensionalization for that problem. We now return to non-dimensionalization and what is the optimal way of doing non-dimensionalization and introducing epsilon into our problem. The important thing to realize is that that the choice of typically for a problem for a mathematical model that we want to set up. In this case, we will look at a simple example now, later we will have to do the same for our interfacial ways problem. Typically there are more than one choice for length time, length and time scales. This leads to some kind of ambiguity for the choice of, for the correct choice of non-dimensionalization for quantities like velocity and so on. Here we have to use our physical intuition in order to choose the scales correctly. Let us look at an example. So let us take a simple example where let us say we are standing on the surface of earth and we are throwing a ball upwards at a speed v0. Now of course the distance to which the ball travels will depend on the speed v0. We typically if we throw with our hands we cannot throw the ball very high. When I say very high I mean the distance that it would travel would be very small compared to for example the radius of the earth. Let us write down some numbers to form our estimates. So typically v0 I will take an upper limit on what is the fastest that a human hand can throw a ball. So typically a fast bowler for example in cricket can throw typically at about 130 kilometers per hour. This is approximately it is slightly more than this but approximately this is 35 meter per second. It is about 36 point something but I have rounded it off to 35. So this is a representative estimate of how fast can the human hand throw. So if we have a fast bowler in cricket throwing the ball is expected to be thrown at a typical velocity of 35 meter per second. Now you can see that this is much much less than for example the escape velocity. We all know how to estimate escape velocity recall that the escape velocity is given by the formula 2 gm by r0. Here g is the universal gravitational constant appearing in the law of gravitation m is the mass of earth and r0 is the radius of earth. Now we know that when you throw a ball upwards the acceleration the force that the earth exerts on the ball via gravity is not really a constant. It depends on the separation between the ball and the earth and the center of the earth. Now typically for the speeds at which we throw balls the distance at which it travels is so small compared to the radius of the earth that as a first approximation it is usually safe to ignore that distance and treat the force as if it is a constant. That gives us a constant value of force consequently a constant value of acceleration which we know to be small g which is 9.8 meter per second square. Now suppose instead of throwing 130 at 130 kilometers per hour we throw at a much higher speed you know 10 times or maybe even 100 times of that. So this is a hypothetical situation where we are launching the ball not by hand but using some kind of a mechanism which can let us say launch it at about instead of 130 kilometers per hour we are launching it at let us say 10 times. Let us compare it with the escape velocity. If you estimate this so g is 6.67 so these are the numbers in SI units. If you calculate if you plug in it here you will get approximately 11 kilometer per second. So you can see that this is even if I launch it at 1300 kilometers per hour that is much much less than 11 kilometers per second. So you can see that 11 to 3600 will be much more than that. So now compare a hypothetical situation where I am launching it at a high speed the speed is high enough such that I do not want to ignore the fact that the as the ball goes upwards the force that the earth exerts on the ball actually changes. So I want to solve the exact equations without making an approximation that the acceleration or the force is a constant. We know how to do this. So let us do this. So the exact force so the exact expression would be so let us say that this direction is x we are solving a one-dimensional problem. So mass into acceleration because the force is always attractive it will be directed towards the center of the earth and we take all quantities in the radially outward direction to be positive so upward direction is positive. So the force is negative and by Newton's law of gravitation the force is just this. Here small m is the mass of the body that I am launching. Now this cancels out and so we get x double dot is equal to minus gm by r plus x square. Now you can immediately see that for the typical velocities which so suppose we will launch it at 35 meter per second or 130 kilometers per hour it would the distance that x max the distance that the ball would travel would be far, far less than r. And so when we solve this equation as a first approximation I can just ignore the x in the denominator and just replace this as minus gm by r square. This is essentially minus gm by r square is g which is 9.8 meter per second square. Now suppose I do not want to do that approximation because I want to throw the ball at increasingly higher and higher speeds. The speeds are sufficiently large that the ball goes a distance but it is still a much bigger distance than what it would if I launched it with 35 but they are still small compared to the radius of the earth and the velocity is still much smaller than the escape velocity. So let us see if we can systematically set up a way of non-dimensionalizing this problem choosing the scales correctly. So how do I choose this? So I want to solve the full problem. I do not want to make this approximation. This is just an exercise in non-dimensionalization followed by solving this problem using perturbative techniques. So I do not want to make this approximation to start with. However I also have an intuition that if my v0 is sufficiently small then at the lowest order of approximation this should be my this problem x double dot is equal to minus g should be my first approximation. So with that in mind let us state the problem. So x double dot is d square x by dt square is equal to minus gm by r plus x square. Now I will launch it at time t equal to 0 it was at 0 my origin of my coordinate system is here. This is x in the vertical direction and I am launching it with some velocity v0. Now let us say that I am I want to write it like this so I pull an r out so it will come out as r square and then I will have x by r whole square. I am doing this because I want to write minus gm by r square as a single variable which is g small g. So this is minus g by 1 plus x by r whole square. This is a very useful form because it tells us that if x max is less than r then a first approximation is just x double dot is equal to minus g. So the acceleration is a constant. Now I let us let us try to first non-dimensionalize this problem appropriately. So if I launch the my projectile with the speed v0 I expect it to travel expected to travel a distance v0 square by g. You can get this from a simple estimate assume that acceleration is constant and if you launch something with speed v0 then how much is the distance traveled in some time t. So this is the maximum distance that it would travel before which after which it would start falling. Now it would travel this distance time taken would be if I know my distance and if I divide by the speed with which it travels then I get a time. So that is v0 by g. So this is my choice of length scale this is my choice of time scale this is something v0 by g some quantity which has the units of time v0 square by g some quantity which has the units of length. We are going to non-dimensionalize by these scales my governing equation. Note that the choice of distance and the choice of time is not unique. If I had gone back to my equation and if I had not done this step where I have replaced gm by r square then I had more dimensional quantities like g capital M and r square. In particular I could have non-dimensionalized by my x with the radius of the earth. I encourage you to try that but first let us do it in the intuitively correct way. So I expect it to travel a distance l before it starts falling back I expect to travel that distance in this time which is given by v0 by t. Now let us scale my equations. So I will write it now in terms of d square x by dt square is equal to minus g by 1 plus x by r whole square. This is just the equation that we had written earlier. Now I define a non-dimensional x which is an x with a tilde which is x the dimensional x divided by something with the units of with the dimensions of length. I will choose that to be v0 square by g this scale. I also have to define non-dimensionalize time and define a new time variable t tilde it is a non-dimensional variable. So dimensional time divided by something with the dimensions of time which is this. Now using these scales I have to non-dimensionalize my governing equation. So governing, notice that this is a non-linear equation. If you solve the full problem it is a non-linear equation in x may or may not be solvable analytically but we are going to try to solve this using a perturbative procedure. So first let us non-dimensionalize it. So if you, so I will have a v0 square by g in the numerator and I will have v0 square by g square in the denominator and then this will become d square x tilde by dt tilde square. And then I will have minus g 1 plus. You can see that I will have a v0 square by rg into x tilde. You can see that this quantity v0 square by rg is non-dimensional. This is length square by time square. So this is a non-dimensional number. I will call it epsilon. What we have on the left hand side after cancellation you can see is just g d square x tilde by dt tilde square and then on the right hand side minus g by 1 plus epsilon x tilde whole square. I can cancel out the g. What about my initial conditions? My initial conditions is x tilde of 0 is anyway 0, dx tilde by dt tilde at time t equal to 0 would be 1. So I have to solve this equation with those initial conditions. Let us estimate the value of epsilon v0 square by rg. And you can see that if I use the formula for small g that I had given earlier then small g was so it is 1 by g. So it will be r square by gm and this is equal to v0 square divided by gm by r. Now this is equal to v0 by square root gm by r whole square. What is square root gm by r? We have seen that 2 times square root v of gm by r is the escape velocity. Square root of 2 times gm by r is the escape velocity. So this epsilon is nothing but v escape. We have seen that even if you make v0 as instead of 135 kilometers per hour we even if you make it 1350 10 times if I make it instead of 130 I make it 1300 kilometers per hour. So it will travel 1300 kilometers in an hour. This travels 11 kilometer in a second. So in 3600 seconds which is an hour it will travel much much more than 1300. So even if you make it 10 times this ratio v0 square v0 by v escape is much much less than 1. However if you throw it at sufficiently large v0 those corrections to the gravitational force might become important. So let us find out a way to solve the full problem in a perturbative manner assuming that epsilon is less than 1 is much much less than 1 and so we can solve it perturbatively but epsilon may not be so small that we can consider force to be a constant. So our scaled problem looks like this. How do we know that this is the correct way of non-dimensionalizing? First check whether when epsilon goes to 0 do you recover the correct equation or not that you would expect on physical grounds. So if you said epsilon equal to 0 you get the equation d square your unperturbed problem is just this. This is just equivalent to saying that if my epsilon is extremely small so in the limit of epsilon going rather if you think of it like this in the limit of epsilon being extremely small I just recover the fact that force is a constant and acceleration is just g. This is just the scale version of that. However as if epsilon is small but not very small then I can use a perturbative procedure to on this equation. I encourage you to try this non-dimensionalization by choosing other scales. So for example we have chosen the length the length scale to be v0 square by g you could have chosen the length scale to be the radius of the earth. You could have chosen the time scale to be the time taken to cover the radius of the earth at a speed v0. Try choosing those scales and non-dimensionalizing your problem and see what happens to the problem in the limit of epsilon going to 0. You will typically find that it will reduce to either something which does not make mathematical sense or it violates your basic intuition that for sufficiently small epsilon taking acceleration to be constant is a valid approximation. So now this non-dimensionalization has added a non-dimensional parameter to our problem which is epsilon. In this problem it is the square of the speed with which we are launching our projectile to the escape velocity of the earth. Provided we do not launch our projectile at the escape velocity epsilon is usually going to be much lower than 1. Let us solve this problem perturbatively and find out what does the perturbed solution look like. So just like before we think of this as an order epsilon term because x tilde is an order 1 term. So we are saying that x tilde is of order 1 after non-dimensionalization if we have non-dimensionalized correctly and so this whole thing is an order epsilon term. So and the left hand side is again order 1 because there is no epsilon anywhere. So let us say that we are going to set up a perturbation approximation to the solution which will be x0 tilde which will be your function of t tilde plus epsilon x1 tilde plus epsilon square x2 tilde plus dot dot dot. We are going to substitute this into our equation and initial conditions and then work out the solutions to this problem at every order of epsilon. Once we get that then we will have to solve the differential equation at every order. An important thing to notice is that this procedure will reduce a non-linear differential equation into a series of linear but inhomogeneous differential equations. It will be easier to solve those equations because they are linear. However the right hand side of the equation the inhomogeneity will become more and more complicated as we go to higher orders. Let us do this exercise and understand how it works. We are going to first let us work on the initial conditions first. So our initial conditions where so let me write down the expansion again. So x of t tilde is x0 tilde. This is my base state solution. I expect the base state solution to give me the lowest order approximation as epsilon goes to 0. The lowest order is just force is equal to constant. So it should just tell me that d square x0 by dt square is equal to minus 1 plus now my initial condition is this. So what does it translate to? So we will have to use this to obtain initial conditions for every order. So how do we do this? So we say x0 plus epsilon x1 of 0 square is equal to 0. This is obtained by substituting the above into this equation. We also obtain at 0. I can satisfy the first equation by choosing. I am equating the coefficient of every power of epsilon to be 0. So I will have and so on. I have to for the second one I have to be a little bit more careful. I there is an order one term here. This is an order epsilon. This is an order epsilon square. So if I collect all the order one terms which are basically coefficients of epsilon to the power 0 then I get. So this is this is all let me write it at the same horizontal level. So this is order one, this and this. Then at order epsilon we will have then at order epsilon square we will have. And like before if I shift the 1 to the left hand side here then I have 0 on the right hand side. So I have to set all the coefficients to 0 in order to satisfy. So this is my initial condition. These are my initial conditions at order one. This is my initial condition at order epsilon. This is my initial condition at order epsilon square. So you can see that we have 0 0 initial conditions at every order except one except at the lowest order. So this will particularly simplify the algebra. So now what we have to do is we have to go back and substitute this form into our governing differential equations. Let us do that. So if I substitute then the left hand side becomes the left hand side of my equation becomes d square x naught tilde by dt tilde square plus epsilon d square x1 tilde plus dot dot dot. The right hand side has to be worked on. The right hand side is 1 plus x tilde whole square which is or epsilon x tilde whole square. So this is epsilon into x naught tilde plus epsilon x1. And now I will have to bring it up in the numerator using binomial expansion. So this just becomes 1 plus epsilon the same thing to the power minus 2. And if you expand it you are going to get. So the exponent comes as a coefficient and then we just have this and then there will be more terms in the expansion. So now I have to set left hand side is equal to right hand side. Collect all terms which are order 1 that will give me my equation governing x0 tilde. Collect all terms order epsilon that will give me an equation governing x1 tilde. In this equation you will find that x0 tilde appears on the right hand side. We cannot solve the order epsilon equation unless we have solved the order 1 equation. This is also true at every order. At every order we can only solve it at that order provided we have solved at all previous orders. In the next video we will do this, we will continue this. We will write down the equations obtained at every order, pair them with the initial conditions that I wrote in the previous slide and then solve them individually. You will see that at every order we will get a linear differential equation. But it will become inhomogeneous as we go to higher and higher orders. And that will give us once we solve those equations it will recover our part of the solution to the non-linear ordinary differential equation.