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Now what we are going to do is to solve the 2D burgers equation before we actually start solving it.

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Let's look a little bit on the equation and just understand the terms that we need to consider.

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Of course, this time the solution will be.

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The numerically finite difference, however.

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In later stages.

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We will solve the using pens.

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But first let's try deal with a 2D equation.

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So in here what we can see is we have two equations actually.

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The first one is du dt, which is basically the change, the temporal difference, the change of U in

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relation to temperature to sorry, the time and the other term is multiplied u d u d which is the spatial.

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Well, the spatial component in which you will change with x plus v, d, u d y, in which also u will

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change with y.

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So this is the first one is in the horizontal direction and the second equation is in the vertical direction.

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So we call it u v.

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Now, after that, we have new this one, which is the diffusion coefficient.

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The kinematic diffusion coefficient.

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The main thing like it's just shows how strong this diffusion or we can control how strong the diffusion

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is from this factor.

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And so after that we have it will be multiplied by a.

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The second derivative of U with X and the second derivative of U with Y.

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So basically, this is the the the U in the X direction, and the other one is the V, which is the

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velocity in the Y direction.

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And we can see it in here.

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Basically it's the same thing like DV become DV over dt du over d x become dv over d dx plus dv over

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dy.

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Just look into or like notice this u is still here.

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Become here and here is going to be the same.

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This is going to be same.

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The only thing will change is this component.

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And it goes like all the way.

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Basically the same thing in the end is everything will change with v over d, v over d.

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The diffusion coefficient Nu is considered to be like similar in both x and y direction.

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If we don't want to do that, we have to, you know, like we have a coefficient for this and coefficient

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for this.

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If this is the same then it will be we just put it here.

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Okay, so this is how the burgers equation are going to work and the components of the burgers equation.

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How can we make it into a code?

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So in order to do so, let's write a code related with burgers equations.

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We can start by writing.

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The name of this file is Burgers.

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Uh, 2D and Main.

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I like to call Main because it is always.

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Because I have many files.

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It's and it's kind of nice thing to have in programming main to know which is the main file.

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So import also like we call the now the first thing is we call the A libraries.

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We need numpy.

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As imp import.

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Matplot.

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Lip.

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Dot pyplot as plt import seaborn.

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Although I don't think we will use Seaborn anyway.

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It's nice.

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And shift enter.

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So after that, we need to start.

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Uh, well, the post-processing we need to set the problem that we have.

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And to do so, we have to consider a n t, which is the number of time step we can consider.

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N t equals 500 and x equals 51 and y equals 51.

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And C equals one.

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And she will be.

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I think it's.

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No, we don't need this.

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This is not needed.

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And u equals a point one.

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And d t equals because we don't have C, we have new the diffusive term and actually we're not really

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using two diffusion like in different directions.

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So this is why it's I got confused a little bit.

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So here we have in new is 0.1 and 0.001 for the time step.

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So this is going to be time step and this is 500.

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So 500 multiplied.

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This one, it will be well, just.

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A 0.5 second.

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0.5 second.

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Yeah.

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So so it will it's not so much time for this, but of course, this is kind of we're not really care

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about the dimensions, we're just numerically solving the things.

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But of course, when you deal with a real numerical problem, we have to be very careful about the dimensions

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and the common sense of what we are calculating.

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But as we learned the numerical things, it's okay, we just write something and we solve the equation.

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So D, x A and x minus one.

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Divided by two because our domain is going to be from 0 to 2, and in Y, it's going to be another from

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0 to 2 and again in X and Y minus one divided by the whole domain like length.

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And now X.

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We start, we will start making the domain we have.

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So the first thing is we need a Or we can just put it here like this and shift enter and then we start

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defining the domain.

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So np dot line space and from 0 to 2 and we have an x steps and x steps 51 means we have actually 50

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until 50 steps.

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So here we have the line step and equals.

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Here in.

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Dot line space also from 0 to 2 and in Y.

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Yep.

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So.

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This is a.

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Is just going to to make a basically a list, a list like this will show that will have these line spaces

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starting from 0 to 2.

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Now, what we need to do is we need to make a grid.

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So in order to to do that, like basically 51 by 51 a grid, we.

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This is going to be our grid.

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And we can initiate it.

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Zero.

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In X and in Y.

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Shift into an X0.

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Yeah.

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So now what we have is.

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A we have a 51 rows and every row will have 51 components so we can see this shape.

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Shape.

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Is 51 by 51.

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So now we have a rather than.

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Only like a line.

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In both cases we will have now a mesh.

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Of course, this is just a placeholder.

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It's not going.

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We didn't really populate it with actual numbers.

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Of course, it's the type of it is numpy array, as we already use numpy.

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So now what we need to do is we need to a placeholder for everything else.

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And the reason I'm doing this because we have to be careful when we are making the placeholders and

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when we are populating them.

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So here you in sorry, in X or in y.

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In x in no problem.

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Like we can we can do it.

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And here.

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Actually this one, the sign convention in this code is in Y in X.

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This is better in Y.

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In x.

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Anyway, it's not going to be different.

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They are the same, but just in case.

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So you.

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We initiated U and V, and this is the going to be the speed at the time step we are dealing with.

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But we need a placeholder for the time step one time step before because the method we are using is

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explicit method.

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We're actually calculating every time step.

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So we take the time step and we calculate the the next time step.

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The scheme is going to be central difference scheme and basically we will compute everything in terms

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of the previous time and we will update the solution for the next time step.

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So this is called a explicit.

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The other option is called implicit in which we consider the time is another vector and we solve all

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the time steps together by converging into, we call it implicit methods.

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It's another we have iterations that we need kind of like to reduce the loss.

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So.

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Yeah, this is basically what we are doing.

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What we are doing.

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Just just a side note.

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So u v u n which is the previous, um, space holder.

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We keep it here and.

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Also you.

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V in which is the previous space, is also here.

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F is going to be the actual.

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So this is the time step, the next time step or yeah, the next time step.

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This is the previous time step and it's or this is yeah this is the the current time step, the one

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we are calculating this is the one be be before it and this one is going to be the, the one in which

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we will save all our solutions.

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So again, zeros, but we need to consider the time.

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So we say NY and X, so this one we have in T and in T, I think we already identified it.

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500.

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Yes, this one.

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So yeah, this is the.

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Basically the the placeholder for the U and V, the solution at the end.

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Again burgers equation we need to solve u and v, we already have x and y and we need to solve u and

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v.

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So this is u v u.

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The previous time step and the basically the we save this the solution and then shift enter just a problem

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here zero.

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Sorry I always forget to put s.

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Now.

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It's okay if you want to see the shape.

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Of course.

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Shape.

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It will be 51 by 51.

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And this one, the placeholder, it will be a little bit bigger.

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Or like one dimension bigger, which is we will have 500.

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So it will be this array.

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So now.

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Pretty much we are defined our problem.

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But what we need to do is we need initial condition and boundary condition.

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Now, the boundary condition, we will keep updating the boundary, basically not allowing the solver

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to update the boundary points at when we are doing the solving loop.

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So this is where we are going to put the boundary condition.

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But what we can do now is to set the initial conditions.

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To do that, we start.

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A basically.

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We take.

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We need you.

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V.

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This one, which is the current time and we need.

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Even this one.

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So how we can initiate it?

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It's, um.

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Or let's start by.

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By first populating everything with ones.

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So the reason is what we need to do is we need, um, we need this.

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I'll just put it this.

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This is going to be our initial condition.

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It's the whole domain.

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Will be like this with all the domain will have a speed of one in you and one in V.

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However, we will have a little bit higher speed in here or not a little bit.

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We will put it as five.

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So here we will have five.

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So our domain will will basically look like this.

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Here is one.

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And of course, this is in you.

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I put it.

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You.

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All of this is the direction of speed is one, except this area.

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It will have a high speed, which is like a five.

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So this is what we are going to do.

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So the first thing is we will put everything is one.

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And to do so, we just put np.ones.

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So this is our current time and here we will put everything.

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So even all the time steps will be one.

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Or you can just put this one to be zero or just we populate specific like specific, uh, the just only

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the zero, the initial condition.

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But anyway, we will update everything and here it's, it's easier to do it like this, although we

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are rewriting this component, but it's not a problem.

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The reason I did like this is to show that there are we we always start with putting the place and later

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we have to identify the initial condition.

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So this is but of course, if you did like this, you just all of them ones you can put this and this

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ones from the beginning.

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But yeah, it's it's it's up to you.

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So the first thing is we put it everything is ones and what we need is we need to, we need to to have

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the you.

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From Integer.

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0.75 over.

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D.

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Basically the location is going to be from 0.7 5 to 1.1.

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.25.

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This is where the location of the high speed.

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And integer.

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1.25.

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We divide by D plus one because the way the python works, it will start from zero.

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So we need to accommodate this one and everything has to be of course, integer So all of this location.

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And.

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The same thing with.

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The other direction.

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So this is the X and this is the Y.

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What is going to be this one is going to be.

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A five.

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Same thing here.

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The.

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X, Y, and it's good.

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The last thing we need.

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And.

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But.

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Oh, sorry.

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We need here.

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We need to add only.

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Zero.

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And shift enter.

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So what we did is we first assign everything to be one.

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And then the initial condition, which is this is our current because we'll start calculating from the

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first one.

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We put it as like basically the initial condition.

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And here the computation for this thing, like for the, the placeholder for um, for our whole computation,

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we set it to be this much of course another way, like we don't really care because we will rewrite

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everything we can just put it like this or sorry, like.

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Like this.

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Like this.

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And it will work because we now overwrite everything all the solution to be initial condition.

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But as we said this one and will be updated later.

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So you do whatever you want.

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I will just put it in a simple way like zero, but it will work.

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Like wherever it is, it will.

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It will it will work just fine.

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Last, let's see our initial condition.

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So we just want to plot the thing x y equals numpy.

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Dot, mish mish.

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I just want to take some space here.

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Mish grid.

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Great.

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And we have X and Y.

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Just a little here.

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And plt dot.

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Figures.

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So this is just a little bit, um, boring thing.

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Like we just need to.

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The plot things.

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But anyway, like it's a course for all levels.

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So please.

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Consider some other people might find it easier to actually we explain everything.

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So we have the contour.

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Contour equals plt dot.

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Contour.

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If.

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X.

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Why?

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And.

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You.

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Everything.

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Everything.

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A map is, of course, many ways to show the the numbers or like the colors.

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Peel.

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This is Jet.

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It is the one.

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We usually like it as engineers.

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PLT title.

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The scientists.

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Of course you solution.

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Plot X.

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Label.

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And this is going to be X and Y label will be Y.

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Just we add lasting.

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Colors bar equals BLT.

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We need to add the legend.

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00:21:33,090 --> 00:21:35,160
Colour bar.

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00:21:38,310 --> 00:21:38,820
Gunter.

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00:21:40,920 --> 00:21:41,580
This code.

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00:21:41,580 --> 00:21:44,760
Last code is just to add the legend.

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Label.

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I don't know where we call it.

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You skill.

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Contours, Matlab.

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Colors.

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00:22:26,010 --> 00:22:26,760
Another buck.

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00:22:30,010 --> 00:22:30,520
Okay.

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00:22:30,520 --> 00:22:32,410
So this is.

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00:22:32,410 --> 00:22:33,400
You have it.

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00:22:33,430 --> 00:22:37,060
Everything is here is one you can see here.

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00:22:37,060 --> 00:22:42,420
And this area is going to be the high speed part.

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00:22:42,430 --> 00:22:45,160
So you is going to be five.

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00:22:46,260 --> 00:22:48,630
Same thing to do to see the.

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00:22:48,810 --> 00:22:50,880
Of course, this is only with you.

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00:22:50,910 --> 00:22:59,550
What we need also to see is the V, so we just change V here and just write it nicely.

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00:22:59,550 --> 00:23:06,570
V Here your solution becomes v solution small little v we used and.

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00:23:07,520 --> 00:23:11,600
So you this is better use Molotov just to be.

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00:23:13,960 --> 00:23:21,760
And this is you scale, this is you and this is now the scale and this is the solution.

325
00:23:21,760 --> 00:23:25,840
So this is the the direction.

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00:23:25,840 --> 00:23:32,650
So the of course, the speed means like it will go in the X direction one and Y direction one.

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00:23:32,680 --> 00:23:36,100
That means the the total velocity will go like this.

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00:23:36,130 --> 00:23:42,370
Of course, if you want to just you calculate the velocity from this from these two components.

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00:23:42,610 --> 00:23:56,770
So this way we already finished the first part of our class, which is mainly writing post-processing

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the 2D burgers equation and next step we need to solve it.

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And after that of course, the post-processing part.