 So, we saw demos of Laplace's equation last time around and what we will, there are a few things that I left out that I did not mention which I will mention now. Then we will look at applying boundary conditions, I do not know if you remember before the demonstration I was talking about applying various types of boundary conditions, I talked about different problems. So, we will just quickly look at applying what are called Neumann boundary conditions or derivative boundary conditions. We will spend a little time on that and if there are no questions after that we will try to see where we will leave Laplace's equation behind for now, we will come back to it later and go on to other simple problems, fine. So, this sort of came under the, I do not, if you remember this came under the category of some simple problems that is right now just for you to recollect we have shown that we can represent numbers on the computer, arrays on the computer, functions on the computer, derivatives on the computer and having done that we said let us look at a few simple problems where these can be used so that we can represent differential equations on the computer and see if we can solve them. And the first simple problem that we looked at was Laplace's equation, is that fine, right. So, what we will do today, let me just first add a little, so to yesterday's, so we are solving nabla square phi equals 0 and the discrete version of that, right, we did the discrete version of that as I indicated earlier, so if phi h is the solution, right, it is a candidate solution, right or if you want to deal directly with the continuous equation, if phi is the candidate solution, this little phi is the actual solution, so if you were to substitute this in there, right, if it were a solution then del squared of this, nabla squared of this would be 0, okay. But if it is not then it will leave a residue that is you will get nabla squared phi equals some residue, okay, you can call it r, so in fact in this case in two dimensions it would be r of x, y, fine. So yesterday in the demo that one of the reasons why I did that 5 by 5 matrix again was I wanted to actually show that one was to calculate the difference between two iterates and the other was to use the residue, okay and I would suggest that you do both without adding too much detail, right, I would suggest that you do both, especially in SOR you may find that there is an interesting difference, okay, especially in SOR you may find that there is an interesting difference. So I would suggest that you would find both of these. Now the residue of course if you have a domain, we have been taking a square domain so far, so the residue r x y is defined within the domain, right, on the boundary capital phi and little phi satisfy the boundary condition therefore it is 0, the residue is 0, okay, the residue is non-zero. So if you want to take a grid on your grid, on your grid at each point you will actually have an Rij, right, at each point you will actually have an Rij, is that fine, which is the residue at that point which you can find by taking Laplace's equation that is the average of these 4 quantities minus 4 times the middle quantity, right and if that is not 0 then you have a residue at that point, fine, okay and if you say that I want to find out what is the magnitude of the residue, you have to use the dot product again, okay. So the dot product we defined so the dot product of r, r would be nothing but the integral over whatever the domain is, in this case this is the domain, okay, r squared say dS or whatever, am I making sense, okay, I just very quietly extended the definition that we gave earlier, I just said by making it d, okay, the domain of definition and it works, it will work. So in reality you try this out, you repeat the programs, whatever I wrote you repeat them and you will see that the value of the error that I plotted and the value of the residue that you get, the error that I plotted that seems to depend on the grid size, actually the way we calculated it could depend on the grid size, whereas if you do it this way it is not going to depend on the grid size, okay, this is actually an integral, this is actually an integral. So if you were to, right, so you please, you want to evaluate this or you want me to write the expression to evaluate this, this is double integral, this is an integral over an area, okay. So you can try to evaluate it and see what you get, okay. So when I say the norm of r, when I say the norm of r, I mean the square root of this, I mean the square root of that and in a sense it will turn out to be the square root of the sums of ri, j, so on, right, okay, but you have to be a bit careful because when you evaluate these integrals, there is the, you can use, when you evaluate the integral for instance you can take ri, j times this area to find the integral over that area, something of that sort, okay, is that fine, okay, only thing is it is r squared here, it is not ri, j, that is the thing to remember, okay, is that fine, everybody is with me, okay. The second thing is, before we go on and since I have talked about the residue, an apt point here is, we have so far taken the actual solution to be x squared – y squared, okay, you have solved a problem where we have taken boundary conditions so that the actual solution is x squared – y squared, so it is possible for you to verify that your code works. I chose this because it makes life easy for you, you are writing this program possibly for the first time, right, I wanted to keep everything simple. What was the truncation error for the second derivative term, do you remember, what the truncation error was for the second derivative term? What was the derivative involved? Fourth derivative, the truncation error is like some delta x squared but the derivative was fourth derivative, assuming that we are representing the derivative, I am sorry, derivative is fourth derivative, right, or if you say phi instead of u, derivative is fourth derivative, ever making sense? So the truncation error by this is basically 0, right, so at this point I would suggest that you choose an analytic function where the truncation error is not 0, this derivative is not 0. So either you choose something that has a fifth degree polynomial z power 5, z power 4 something of that sort, z is a complex number, either choose the real of z power 5, do something like this, that is one possibility, right, or pick something like sin z, okay, pick a transcendental function, okay, exponential of z, hyperbolic tangent of z, something of that sort, right, okay, fine. And see how well your code works, right, so far the problem that I have given you unit square, everything is well behaved, it was deliberately chosen. Now you have a situation where you are going to pick a function which is where the truncation error will not automatically be 0, okay, so see how well it behaves, right, how well does the code behave as well when the truncation error is there any change in behavior, so just try it out, okay, right, so this is important. So the process that we are going through right now is also important, so anytime you encounter the equation for the first time, right, if you or a class of equations you should see whether there is a sub-problem for which you have an analytic solution, pick a simple analytic solution and test your code against that analytic solution, okay. Then you can make the known solution as more complicated, do not try to do it all in one shot, that is not a good idea, okay, do not try to write the program all in one shot, is that fine. The second thing is what if you have an equation for which you do not have a solution, you do not have an analytic solution, right, where here is supposed to be computation of fluid dynamics, so we are not going to look at Navier-Stokes equations in great detail, right, right, it is an introduction to computation of fluid dynamics, so we are really not going to get to Navier-Stokes equations directly, but there are not that many analytic solutions to Navier-Stokes equations, so what do you do if you do not have an analytic solution, how do you test your code, how do you test your program, so one way to do it would be let us take this, let us take Laplacian of V equals 0 and say that we do not know of a way to generate a solution for this on a unit square, okay, so then you just guess something, so just say instead of saying x squared-y squared we guessed x squared-y squared, we guess x squared-y squared, so anyway you know it is not a solution already, right, so if you substitute it, it will give you a residue, what does the residue that it gives you, 2 plus 2, right, residue that it gives you as 2 plus 2, so this is a solution to Navier-squared phi equals 4, right, so I started off with an equation for which I did no solution, now I have an equation for which I know the solution, am I making sense, of course there may be difficulties associated, sometimes there are difficulties associated with adding a right hand side, okay, right, but if you are trying to figure out how well am I representing this Navier-squared phi, how well am I representing these derivatives, how well is it working, then it is possible for me to actually turn around, substitute some guess solution into the equation, it leaves a residue and just say well that does it for me, okay, that is an equation for which I have a solution, is that clear, okay, because very often you will hear people saying oh I am going to compare it to experiment because I do not have a solution, I am going to compare it to you know the solution of another code because I do not have a solution, not necessary, right, you can generate an equation that is very close for which you have a solution, what it will do is it will add typically a source term that has its own consequences, that has its own consequences as long as you are aware of that, but basically you can turn around and generate an equation for which you have a solution, is that fine, okay, right, so this is really as far as the residue goes, so the residue is not only useful to check whether you have a solution, the residue is also very useful for you to generate a solution, okay, is that fine, everyone, right, let us do the last part, I was talking about the Neumann conditions, so conditions. So effectively what you can have is, you can have, we still stick with our unit square, okay, there is a very simple reason why I am sticking to a unit square or you can change to a rectangle or whatever, if the geometry change then we cannot use a Cartesian mesh, right, then we would have to do something special, either we would have to use unequal meshes or we would have to get into some, the game of grid generation which is a completely different course, okay, so I am restricting myself to squares and rectangles simply because it illustrates what we need for the introduction to CFD part, so on one of these sides, right, it is possible that what you have is a condition that is like dou phi dou n equals either a function or 0, okay, in particular 0, I say 0 because this is a condition that we are used to include mechanics, right, dou phi dou n equals 0 and on the other sides you may actually be given a function of, you may actually be given functions of x and y, am I making sense, okay, so if you want to discretize this, if you want to break this up using grid lines, okay, on these nodal values it is possible for you to find the phi value directly from the function specified, so what do we do here, what do we do at the bottom, I need a proposal. So I can use a finite difference method to represent, so I can use a finite difference method to represent dou phi dou n at this point, okay and there are many ways by which we can do it, one possibility is that the simplest thing, so if this point is 1, this point is 1, that point is 2, then you can just basically say phi 2 – phi 1 divided by delta y is 0 telling us phi 1, that is easy enough to do, okay, so that is a boundary condition, so you would actually evaluate this point on the boundary, you would actually calculate this point in the boundary, where values of the function are given on the boundary, they remain a constant, they do not change, where the value is not given on the boundary, it becomes part of your iteration, whatever the value is at 2, the same value is given at 1, that is one possibility, okay, is that fine. If you are not satisfied with the truncation error of this representation, you can actually use 3 points, right, you can use a higher order 3-point representation to find the first derivative at this point, okay, everybody, right and at this point, at this juncture, I want to point out something, it is okay that we had dou phi dou n equals 0, you could have dou phi dou n equals some function, either function of n or function of y or whatever it is, you could have dou phi dou n not a function of n, but dou phi dou n is a function, some value, okay, dou phi dou n something that evaluates that changes, okay, that changes one, it could change along this length, right, so in this case, I do not want to say t or whatever, so dou phi dou n which is a function, I just put a dot there and your, the problem may be that if you have to integrate this, right, consider a situation where for example that this is heat flux or something of that sort, so somebody gives you a boundary condition where this is heat flux, they are telling you how much energy is flowing into the system, okay, so this is the rate at which the energy is flowing into the system, this is f, right, so you say what is the big deal, so you just set this derivative equals the f, right, that is what it looks like, in this case, you have to be very careful as to whether what is the nature of this, whether it is a, whether it is a, whether the direction is important, okay, very often we do not think about it, so go back to the discussion that we had, so you have dou phi dou n equals 0, this is x direction and dou phi dou n equals 0 or dou phi dou n equals some function on this boundary, okay, so I am working up to something, I mean you are wondering what is the heat hockey about, see the idea is when you normally talk about a derivative, there are two things that are, the derivative basically has two components to it, right, the way I would like to think of it, it is basically linear transformation in the direction, okay, this is something that I mentioned earlier at the beginning of the semester, so the derivative basically consists of a linear transformation in the direction, it is very important for me as to where I am going to go now, right, and you are used to it from the, you are used to it from the multivariate calculus point of view, you are not quite what you call it, we do not quite think about it when we are talking about calculus of one variable, okay, derivatives of one variable, so you are used to a directional derivative where I say grad phi dotted with n, right, or grad phi dotted with ds, some differential element gives me d phi, you are used to this, right, you are all familiar with this, this is the, so this is the linear transformation, this is the direction, this is really the definition of a derivative, okay, now you say so what, what, how does this, how is this important, well it makes a difference, you can say the derivative is something at some point, but it makes a difference whether you are walking uphill or walking downhill, the direction is important, it is not enough to say that oh I am on a mountain, it is important to you to know whether you are walking uphill or walking downhill, the direction is important, do you understand, so if you are applying a boundary condition which is based on a derivative, most of the time when you, when you say derivative, when you are thinking derivative, when we take a derivative of some function, we are implicitly thinking in terms of positive x direction, do you understand what I am saying, most of the times you are saying dou f, dou x dx is dy, df, implicitly in your mind you are implicitly thinking of dx being a positive quantity, it could be negative, you could be going in the other direction, it could be going downhill, okay, so anytime you are applying derivative boundary conditions, you have to be careful, make sure that you are evaluating the derivative properly, if it is, the derivative equals 0, it is immaterial, but if the derivative equals something, then you have to pay attention, right, especially in multiple dimension there is a direction involved, pay attention to what you are doing, you can get the sign wrong, it is very easy to get the sign wrong, okay, is that fine, are there any questions, right, see normally in derivatives of one dimensions we do not bother, but usually the place where students get into trouble is if they are talking about solving the heat equation which we will look at it a little later in the semester and I give a boundary condition, a derivative boundary condition on the right hand side because I am a mean guy, right, I give a derivative boundary condition on the right hand side which is not, which has a flux term, if it is a vector, then typically people run into difficulty because they do not remember that the derivative actually has a direction associated, it is a linear transformation and a direction and in one dimension we do not think about it, but in multiple dimensions it is very clear that it is, in fact in multiple dimensions very often you call it directional derivative, it is always directional derivative, okay, it just so happens that in one dimension the direction is dx, that is what you are thinking about, okay, is that fine, are there any questions? So what you can do is you can possibly try again Laplace's equation, try to apply Neumann conditions and right, set dou phi dou n equals 0 on one of the sides and see what happens, what happens to you, is that fine, okay, right, so what we will do is we will now change gears as I indicated, so we have looked at, we have looked at this equation so far, let us look at a different equation, we will come back to this later in the semester, let us look at a different equation, I am going to look at the equation dou u dou t, let us say dou u dou x equals 0, is that fine, so this is the first order, simple equation it has a long name, first order linear one-dimensional wave equation, so it says it all, all the derivatives are first derivatives, okay, the equation is a linear equation, you can verify that it is linear and it is one-dimensional in the sense that it is in one-space dimension and I have introduced a new term which is time, okay, actually it is still 2D, as far as we are concerned it is still two-dimensional because it has x and t instead of x and y, I mean I may have well have written dou u dou y plus a dou u dou x equals 0, right, it is just chalk dust, we just interpret that y as t, right, but because it is in one-space dimension, we refer to it as one-dimensional problem, is that fine, okay, so what is the nature of this equation, what is this equation, what is the behavior of this equation, is it possible for us, we had an analytic solution for Laplace's equation, is it possible for us to get an analytic solution to this equation, okay, so you may have seen this, you may have seen this or a variant of this in your partial differential equations course, let me just quickly go through this, it is possible for me to write this in terms as a directional derivative, actually possible for me to write this as a directional derivative, how do I do that, so do not let it bother you that this is t, right, as I said it is just chalk dust, it could be y, right, so this is basically looks like j dou by dou t, that is an operator, plus i a dou by dou x, these are unit vectors, that i is not, it is a unit vector in the standard Cartesian coordinate system acting on u, is that fine, equals 0, what am I doing, and this I can split as j plus i a dotted with j dou by dou t plus i dou by dou x on u equals 0, which is of course sum s dot grad of u equals 0, now you will understand why I made such a song and dance about the directional derivative, right, I need a directional derivative, is that fine, where s is j plus ai or ia, okay, and if the length s, if the coordinate s is along s, then this basically tells us from the definition of the directional derivative, if in fact there is a derivative, this tells us that du ds equals 0 or u is constant along s, is that fine, okay, of course if the differential equation the right hand side had not been 0, the right hand side had been something else, then this would be du ds equals f, and then u will not be constant along s, okay, because the right hand side is 0, it happens that u is constant along s, what does this mean, let us look at this pictorially, so that is x, that is t, so in your course in physics maybe you have heard of world lines or something of that sort, right, so this is basically what we are doing, this is an xt plane and we are drawing world lines, right, that is what I, I want you to, if you have run into this, I want you to just think about, right, if you have run into it in your physics, I want you to just remember world lines, okay, that is fine, so this is x, that is t, this is what we are saying, so let us say that we are solving, we are given that differential equation and we are given a condition along t equals 0, a time t equals 0, we are given what is the state of u, what this basically says is that along the direction s, along the direction in, this is s, this is along the direction s, this is a vector s, right, starting at this point I am going to measure the length s and this basically says along this line du ds equals 0, is that okay, everyone, that means u is constant along this line, so if you take another line and since a is constant, okay, did I mention a is constant, no, but I mentioned that it was linear and if you checked out, if you go back and check to see when is it linear, you would need, right, a to be constant or a function of x, y, it could be a function of x, y, even then it would be linear, okay, so a is a constant, a is a constant, right, a is a function of x, y it would still be linear and I am getting a few shakes, fine, okay, right, so then you would have a different s and even along this line u would be a constant, okay, so we have a differential equation, is there a physical problem for which this differential equation works, so it is very simple, let us just say we have a stream of water, right, so this is the standard example that I give, we have a stream of water that is flowing along the x coordinate direction, okay, so I am here and in that stream of water I am going to add chalk dust, okay, so the chalk dust that was added here at some time t equals 0 travels along the x coordinate direction in time, right, so in x t, so if the stream is moving at a constant speed a, right, after some time, the chalk dust that I added here would have traveled to that point, you understand what I am saying, okay, no, so basically all it is doing is, so here you have a physical problem that is actually represented by this equation, so I have a stream that is moving at a constant speed, I add some marker, some tracer, right, I add some ink dye or I add some chalk dust, I add something and that propagates, the property u that you are talking about is the amount of chalk dust that you added, right, at the type of chalk dust that you added, so I have added blue chalk dust in one spot, I add white chalk dust here and I add blue chalk dust somewhere else and that blue chalk dust travels along that line, okay, at the speed what is the speed, it travels the distance a in unit time, right, so that is 1j, that is 1j and this is ai, it travels the distance a in unit time, is that okay, everyone, okay, so it is very clear, so what this basically does is, what this basically does is, right, it propagates whatever that is there at this given point, at this initial point, it propagates it at the speed a, right, it propagates it at the speed a and as a consequence because du, ds equals 0, which means that it is not as though white chalk dust is being added everywhere, see it is being added only at one point, if I were to along the length of the, along the length if I were to add white chalk dust, right, then this right hand side would be an f, the rate at which I am adding that white chalk dust, okay and as you go along you would be accumulating chalk dust as you go along, fine, another possible example something that may be you are familiar with is if you turn on the water heater in the morning, you want to take a shower, right, very soon you learn that you should not get under the shower and turn on the light tap or faucet or whatever it is because what you will get initially is cold water, it takes time for the water, hot water to travel from the water heater to where you are, right, so if you open the tap to a certain extent if the water is travelling, so there is a front, so you have a pipe, there is hot water here on the left hand side and there is cold water, so you open the tap, this surface, call it a contact surface if you want, this surface propagates at a certain speed, so if it travels at the speed of air then you will know that your cold water and hot water that interface, what is the, what is the along the length of the pipe at each time, at each time instant along the length of the pipe where it is, is that fine, so it is a very simple equation, it is a very simple equation, okay but what it does for us is it picks up this property called advection, we use the term advection because for historical reasons convection is already used up by the convective heat transfer people, so it is convection but just so that there is no confusion at future times, we will introduce the term advection, okay, so something is being carried and in this case because du, ds equals 0, something is being carried without change in identity, is that okay, right, some property is being, some property u is being propagated without change in identity, this is very different from Laplace's equation, Laplace's equation was averaging, right, Laplace's equation was averaging whereas what this is basically doing is, this is carrying, carrying, propagating whatever that you have, whatever you have is being propagated in a certain direction, okay, this is very important, it is being propagated in a certain direction, so for us this is very important, these, so you could clearly what we have done in a sense if you think about it, this is like we are saying if the coordinate system were not this xt coordinate system but were actually aligned along this line, then instead of having a partial differential equation we would have an ordinary differential equation, really that is what this says, this basically says that somehow if you had managed to rotate the coordinate system, the physics of the problem that you are solving, this would be, this would just be an ordinary differential equation, right, see in your mind I do not want you to think, that is why I said this is just shocked us, if you go to, if you go to, of course I mean this example may not help everybody but if you go to a library, right, where the staircase is at an angle with respect to the coordinate systems as seen by the streets, so there are two roads, there are two roads, a library or institute library has two roads that flank it, okay, so that could represent your coordinate system but the staircase to the library is that way, okay, so if you have this as x and y coordinates, if you think about it, what is a given stair, what is this given step, along that the height is a constant, that is the property, you understand, so do not think of it as t, it has to be time, right, it just basically says that along this property is a constant, the height is a constant, so you have these stairs and along different, if I know that the stair one step starts somewhere at a certain height, I guarantee that if you travel along that line, that height will be the same, if you travel in some other direction then you are in for a surprise because the height will suddenly abruptly change, am I making sense, so if you travel along that line, right, the height is a constant, these are basically characteristic directions, is it okay, right, in this x, y coordinate system and there is a coordinate system, if you perform a rotation, you can find out what that orientation is, along which the differential equation for that is, it is a constant, height is a constant, am I making sense, dh, ds or whatever it is, it is 0, is that clear, everybody, right, so it need not be, when I say propagation, it is propagation, when I say propagation, it looks like propagation, that is because there is time but it need not always be in time, the other coordinate need not be time, okay, the other coordinate need not be time, is that fine, everyone, okay, so this equation now results in this kind of a scenario, is there a way for us to somehow solve this, is there any other way that we can, is there anything else that we can do, is there any other way that we can solve this equation, there are lots of examples that I mean, going back, there are lots of examples that I can give for, is there any other way that I can solve this equation, you have to go with the notion that, you have to go with the notion that, right, remember what I said earlier, so we are basically trying to integrate this differential equation, okay, and all integration is guessing, so you guess, you substitute into the differential equation and you try to see whether you are able to get a solution to, right, whether the guess is a solution to the given differential equation or not, am I making sense, by substituting into the differential equation and verifying whether it is a solution or not, so normally what you would do is you would just guess, so with the information that we have, the knowledge that we have, is there a way for us to guess, we will see, maybe we will go along a little further and see whether that, so what could be the nature of the function that is propagated, let us try a few functions and then see where that takes us, maybe that will give us an idea as to what is happening, so if my initial condition, a typical boundary condition could be u at, u at t equals 0, right, u of x, t is a function, u of x, 0 equals, it starts with a simple one, is this enough, have I specified everything that you need to solve the problem, a is 1, where do I give a boundary condition, where do I give a boundary condition, no t equals 0, this is the initial condition, so you have to be a bit careful with, so since we are saying time, now we will come up with two different names, so that is the initial condition, we say initial condition and boundary condition, that is initial condition, okay, why x equals 0, because it is propagating from left to right, because it is propagating from left to right, does that make sense, why cannot I prescribe something at x equals l, let us say that I am actually looking at something propagating through a pipe, the length of the pipe is l, you are saying that I should prescribe the condition at x equals 0, right, so you are saying that I should prescribe the condition at x equals 0, right, that is fine, that is correct, why cannot I prescribe it at x equals l, we cannot, we can or we cannot, we can, if a is negative, if a is negative we can, that is called conversation context, that does not make sense, a is 1, we look at, if you flip sign why it happens is a different story, right, you are just basically saying flip the coordinate system around, yeah, if a is negative what is going to happen, the characteristics sign directions will change, the slope of the characteristics will change, that is the key, so you see the thing is, so think about it, I open the tap and I insist that hot water comes, but hot water does not come, you understand, I cannot make it hot water, I cannot open the tap and insist that the water at the exit of the tap be hot, right, the water in the water heater is hot, I can turn on the switch and make the water in the water heater hot, right, I cannot turn on the switch, see unless we do something to this contraption, but right now water heater pipe, tap, right or faucet, whatever, that is all you have, you cannot insist that the water at this end be hot just because you turned on the water heater, what you can do by turning on the water heater is eventually the hot water will come out, eventually, because you turned the water heater, you understand what I am saying, so you can prescribe the condition here, right, in this case, in this particular case you can prescribe the condition here, you cannot assert, the key thing is assert, you cannot assert, so I can put chalk dust here, if I can only put chalk dust at the inlet, I have a stream of water, if I can only put chalk dust here, I cannot put chalk dust here and insist simultaneously that at a given time, right, chalk dust be something else there, it is not possible and once I have put a certain amount of chalk dust here and the chalk dust has traveled, the chalk dust has traveled or it has gone along the length of, the hot water has gone along the length of the pipe, I cannot at that point insist, I cannot assert saying that no, no, the hot water has, there should not be hot water, I want cold water, right, you are now committed, you let the hot water in, it is in the pipe, it cannot go anywhere else, you let water out, you are going to get hot water, right, you cannot assert that it has to go back to being cold water, right, because remember the boundary condition is an assertion, you are going to say this is the value, this is the value of the temperature, this is the amount of chalk dust that is there, okay, so because of the nature, because of the direction in which the, the characteristics are oriented, right, which is, which comes back to what Ashok was saying, if you change the sign of A, then the propagation direction changes, if A is negative, then you are traveling in the negative x direction, okay, then the property is being propagated from, if A happens to be negative, the property is being propagated from a positive x, larger x quantity to a smaller x quantity, okay, in which case then you cannot prescribe the condition on the left hand side, you can only prescribe the condition on the right hand side, am I making sense, okay, right, so what we need is, we therefore need a condition on the left hand side, which could be u equals u of 0, t, so for all time, right, starting at t equals 0, if it were constant, let us say it is 1, keep life easy, then what are you going to get, this is xt, okay, so on the xt plane, of course, we have the same characteristics, but this time they are all 45 degree lines, because A is 1, all the lines are supposed to be parallel, independent of how I have drawn them, 45 degree lines, and so they also come, there is something there, but we have cut our domain there, so there are 45 degree lines coming from the boundary also, yeah please, meaning what, whatever boundary condition that you give, no, you are saying that instead of being on this, if I give it on some curve, is that what you are saying, yeah, u of x0, t equal to something, no, it does not matter, it really does not matter, I am just trying to keep this, in fact this problem that I am talking about right now is rather boring problem, right, nothing exciting is going to happen, this is rather, this is a, this is a, this is a, this problem that I am, specific problem that I am talking about right now is rather boring problem, right, I will admit that, there are, when we come to the numerics, there are interesting things that we are going to see with respect to this equation, right, but I just want to make sure that, I just want to make sure that there are no issues, right, that we are basically all on the same page, and there are certain elements of this physics that are important for example, so it need, it can be, it can be at x0, what you are basically saying is, this particular equation I can actually integrate back in time also, right, in a sense I think, you understand, right, so this particular equation you can actually integrate back in time, so in a sense, if you say that the value of something, if there is some value here, you can go back in time, but that looks very suspiciously like making a as minus a, sort of, right, okay, if you think about, if you look at, it comes back to the same thing, so you can go back in time, this is an equation that you can, you can integrate in time, so you could say that this is like, you are out, you are, you are, you are hunting for somebody, you are looking for somebody, right, and you find a clue, and then you try it or you get an order, right, you are, so you then try to figure out, you try to work back to see from where does this come, right, and of course all, all, all adventure stories that you have read, the children, they are always talking about being upwind or downwind or whatever of the quarry, right, so you, you do not, you want to be downwind of the quarry so that, right, whoever that you are hunting can't smell you, but you can smell them, right, that kind of a thing, so you, or you can, so you have to look, so that is, that is basically a matter of where was this person, right, where, where is this person, so you, if I give you some value on the characteristic here, it is actually possible that you can trace it back, okay, you can go back in time, is that fine, okay, but you could give the value, you could, you could prescribe the value in reality in any point, u of x, 0 or u of 0, t, yeah, because I mean, as I said, it is all just, t and the x part, right now that it is, whether it is time or not, so that is what, that is where we, that is what you have to look at, that you are able to integrate back, what it finally boils down to is, as I said, whatever coordinate system that you give, that you can rotate the coordinate system to align and get an ODE along this line, right, and get an ODE along, and that, that, that is possible is the nature of that equation, okay, it may not always be possible, sometimes there may be some scaling, stretching, there are other, other, other issues that are, that are involved, okay, is that fine, so, right, at the time we are saying around the, along the characteristics, the, yeah, in this case, the solution is propagated, no, no, if you prescribe the condition here, what I am saying is, you can, if you say it, when you say, when I say prescribe, in a sense, in this case, what you say is, I discovered that the value here, if you insist that this is time, you say I discovered that the value here is 1, then you can integrate back and figure out for, along that x, at what point in time was it 1, you understand, if your domain, if your domain is that, the domain of interest is that, you can look at the exit condition, if you say, if you look at the exit condition, for instance, just say your hot water pipe, your pipe, your pipe in your house is 1 meter long, and the water is travelling at 1 meter per second, so you know that if it is, right, 40 degrees, the 50 degrees, the 60 degrees Celsius at the exit right now, then half a second ago, midpoint it was 50 degrees Celsius, half a second ago, midpoint it was 50 degrees Celsius, is that fine, okay, so you can integrate back in time, or you can integrate forward, from that point only, okay, yeah, yeah, yeah, there seems to be a problem at 0, 0, right, so they say, so we will look at it, we will plot this, we will plot this to see what happens, so there is actually a sort of a step kind of a thing at that point, right, I mean just like you have cold water on one side and hot water on the other side, and there is a very sharp interface between them, is that fine, okay, so we will get back to this in the next class, okay, thank you.