WEBVTT 00:00:00.035 --> 00:00:03.228 A plane channel is the simplest wall-bounded turbulent flow we 00:00:03.228 --> 00:00:05.074 can simulate exactly. 00:00:05.271 --> 00:00:08.996 Two parallel walls, a fluid driven along the gap between them, 00:00:09.079 --> 00:00:12.992 and every eddy in that gap resolved on the grid rather than modelled. 00:00:13.789 --> 00:00:17.061 The two directions along the wall are statistically homogeneous, 00:00:17.144 --> 00:00:19.073 so we take them periodic: 00:00:19.176 --> 00:00:22.367 x downstream, and z across the span. 00:00:22.450 --> 00:00:24.422 The remaining direction, y, 00:00:24.505 --> 00:00:29.220 runs across the gap, from minus one at the lower wall to plus one at the upper. 00:00:30.133 --> 00:00:32.525 At each wall the fluid sticks: 00:00:32.722 --> 00:00:35.602 all three velocity components vanish there. 00:00:35.705 --> 00:00:37.108 That single condition, 00:00:37.191 --> 00:00:41.952 no-slip on a solid boundary, is the source of every difficulty in this lecture. 00:00:42.656 --> 00:00:46.117 Before deriving anything, let me put the destination on the board. 00:00:46.430 --> 00:00:50.471 Kim, Moin and Moser march two scalar fields. 00:00:51.085 --> 00:00:54.079 The first is the velocity component normal to the walls, 00:00:54.162 --> 00:00:56.937 and its equation is fourth order in space. 00:00:57.734 --> 00:01:01.206 The second is the vorticity component normal to the walls. 00:01:01.402 --> 00:01:04.895 Its own equation is a diffusion equation with a source term, 00:01:04.978 --> 00:01:07.440 and it is only second order in space. 00:01:08.144 --> 00:01:09.652 And look at what is missing. 00:01:09.735 --> 00:01:12.022 There is no pressure in either line. 00:01:12.474 --> 00:01:16.246 Two scalars have replaced three velocity components and a pressure, 00:01:16.329 --> 00:01:19.799 and every boundary condition the walls hand us will turn out to be a condition 00:01:19.882 --> 00:01:22.390 on something we are genuinely solving for. 00:01:23.442 --> 00:01:25.963 Here is the system we normally write down. 00:01:26.159 --> 00:01:28.782 Momentum for each of the three velocity components, 00:01:28.865 --> 00:01:33.568 with the pressure gradient sitting in it, and incompressible continuity beside it. 00:01:34.364 --> 00:01:35.526 Count them: 00:01:35.630 --> 00:01:38.196 four unknowns and four equations. 00:01:38.393 --> 00:01:40.611 But look at where the pressure sits. 00:01:40.808 --> 00:01:42.584 It enters only through its gradient 00:01:42.584 --> 00:01:47.101 in the momentum equations, and continuity does not contain it at all. 00:01:47.897 --> 00:01:51.855 Take the divergence of the momentum equations, use continuity, 00:01:51.938 --> 00:01:54.249 and the time derivative drops out. 00:01:54.561 --> 00:01:56.882 What is left is an elliptic problem: 00:01:56.965 --> 00:02:01.737 a Poisson equation whose right hand side is quadratic in the velocity. 00:02:02.441 --> 00:02:04.949 So the pressure is not marched at all. 00:02:05.146 --> 00:02:08.453 At every time step it is determined, everywhere at once, 00:02:08.536 --> 00:02:11.379 by the velocity field everywhere at once. 00:02:11.462 --> 00:02:13.680 And that is where the trouble starts. 00:02:14.477 --> 00:02:17.343 The first difficulty is the one we have just met. 00:02:17.426 --> 00:02:20.547 The pressure has no equation to march, so it has to be recovered 00:02:20.630 --> 00:02:24.706 from an elliptic solve at every step, over the whole channel at once. 00:02:25.410 --> 00:02:29.065 The second is that the velocity has to come out divergence free. 00:02:29.148 --> 00:02:32.502 Solve the momentum equations on their own and it will not. 00:02:32.585 --> 00:02:37.938 So velocity and pressure are two halves of one solve, not two solves in sequence. 00:02:38.734 --> 00:02:43.562 The third is the sharpest. The walls tell us the velocity there: it is zero. 00:02:43.645 --> 00:02:46.107 They tell us nothing at all about the pressure. 00:02:46.559 --> 00:02:50.099 And yet an elliptic problem demands a condition on every boundary, 00:02:50.182 --> 00:02:52.769 so whatever we supply is a numerical invention, 00:02:52.852 --> 00:02:55.755 sitting exactly where the interesting physics lives. 00:02:56.459 --> 00:02:59.975 Kim, Moin and Moser take the obvious way out. 00:03:00.058 --> 00:03:04.146 If the pressure is the problem, remove it from the equations altogether. 00:03:04.249 --> 00:03:06.895 That takes one vector identity and two derivatives, 00:03:06.978 --> 00:03:08.952 and it is the next thing we do. 00:03:09.998 --> 00:03:12.469 Vorticity is the curl of the velocity, 00:03:12.552 --> 00:03:16.790 and the component we are going to need points along the wall-normal direction. 00:03:17.103 --> 00:03:22.015 Here is a wall-parallel plane, with the horizontal velocity drawn on it. 00:03:22.812 --> 00:03:25.097 This particular field circulates. 00:03:25.180 --> 00:03:28.825 It turns about the wall-normal axis, which is perpendicular 00:03:28.825 --> 00:03:34.085 to the plane you are looking at, and that turning is exactly what omega y measures. 00:03:34.998 --> 00:03:39.318 In components, only the two wall-parallel derivatives appear: 00:03:39.514 --> 00:03:43.390 the rate at which the downstream velocity varies across the span, 00:03:43.473 --> 00:03:47.642 minus the rate at which the spanwise velocity varies downstream. 00:03:48.554 --> 00:03:50.297 Now the derivation. 00:03:50.493 --> 00:03:54.625 Collect every nonlinear term into one symbol, N, 00:03:54.708 --> 00:03:57.170 so that the momentum equation reads: 00:03:57.273 --> 00:04:02.928 rate of change equals N, minus the pressure gradient, plus viscous diffusion. 00:04:03.841 --> 00:04:05.955 Take the curl of that equation. 00:04:06.059 --> 00:04:07.614 The pressure term is the gradient 00:04:07.614 --> 00:04:13.478 of a scalar, and the curl of a gradient is identically zero, so it simply disappears. 00:04:13.791 --> 00:04:18.668 That is the whole trick, and the rest of this lecture is bookkeeping around it. 00:04:19.372 --> 00:04:24.920 What comes back is the vorticity transport equation. Take its wall-normal component 00:04:25.004 --> 00:04:28.916 and the second of our two governing equations is already there: 00:04:29.020 --> 00:04:30.818 omega y diffuses, 00:04:30.901 --> 00:04:35.592 and it is stirred by a source built out of the nonlinear terms alone. 00:04:36.389 --> 00:04:39.141 The velocity equation takes more work. 00:04:39.245 --> 00:04:43.204 Start from the same momentum equation and take its divergence. 00:04:43.517 --> 00:04:48.127 Continuity kills the time derivative, and what is left is a Poisson equation 00:04:48.127 --> 00:04:52.528 for the pressure with a right hand side made only of velocities. 00:04:53.928 --> 00:04:55.809 Now take the wall-normal component 00:04:55.809 --> 00:05:00.024 of the momentum equation and hit the whole line with a Laplacian. 00:05:00.220 --> 00:05:02.845 Three of the four terms are harmless. 00:05:03.042 --> 00:05:08.778 The pressure term becomes the Laplacian of a y derivative, which is the y derivative 00:05:08.778 --> 00:05:13.712 of a Laplacian, and we have just worked out what the Laplacian of the pressure is. 00:05:14.509 --> 00:05:17.796 Substitute it in and the pressure is gone for good. 00:05:18.410 --> 00:05:21.848 Rearranged, this is the fourth-order equation: 00:05:22.044 --> 00:05:25.899 the Laplacian of v is what evolves, viscosity acts 00:05:25.899 --> 00:05:31.160 through a fourth derivative, and the source h collects the nonlinear terms. 00:05:32.071 --> 00:05:35.068 There is a neater way to say what we just did. 00:05:35.264 --> 00:05:41.427 Minus the Laplacian of a divergence-free velocity is the curl of its vorticity, 00:05:41.510 --> 00:05:42.822 so applying the Laplacian 00:05:42.822 --> 00:05:47.397 to the wall-normal momentum equation is taking the curl twice. 00:05:47.849 --> 00:05:51.518 The fourth-order operator is what a double curl looks like. 00:05:52.315 --> 00:05:54.347 So here they are together. 00:05:54.544 --> 00:05:58.014 A fourth-order equation for the wall-normal velocity, 00:05:58.097 --> 00:06:01.847 and a second-order equation for the wall-normal vorticity. 00:06:02.644 --> 00:06:06.032 They are coupled, but only through the source terms h, 00:06:06.116 --> 00:06:10.935 which are quadratic in the velocity and will be evaluated explicitly. 00:06:11.131 --> 00:06:15.208 The linear operators on the right are completely independent of each other. 00:06:16.259 --> 00:06:18.173 Fourth order sounds expensive, 00:06:18.256 --> 00:06:22.692 and it is not, because we are about to solve it in one direction only. 00:06:22.888 --> 00:06:25.328 Two questions are still open, though. 00:06:25.431 --> 00:06:28.099 Where did the other two velocity components go, 00:06:28.182 --> 00:06:32.107 and what boundary conditions does a fourth-order equation want? 00:06:32.420 --> 00:06:35.846 Both answers arrive as soon as we go to Fourier space. 00:06:36.939 --> 00:06:40.571 Both wall-parallel directions are homogeneous and periodic, 00:06:40.654 --> 00:06:42.849 so both get Fourier series. 00:06:43.046 --> 00:06:46.564 Write the wall-normal velocity as a sum over modes: 00:06:46.668 --> 00:06:49.372 each mode carries a wavenumber in x, 00:06:49.455 --> 00:06:54.773 a wavenumber in z, and an amplitude that still depends on y and on time. 00:06:55.477 --> 00:07:00.028 Inside a single mode the wall-parallel derivatives are no longer derivatives. 00:07:00.225 --> 00:07:07.225 Differentiating in x multiplies by i k x, differentiating in z multiplies by i k z, 00:07:07.551 --> 00:07:12.625 and the Laplacian collapses to a second derivative in y minus k squared, where k 00:07:12.625 --> 00:07:14.901 squared is the sum of the two squares. 00:07:15.384 --> 00:07:17.983 That is the whole content of the transform, 00:07:18.066 --> 00:07:20.749 and it has already done something drastic. 00:07:20.945 --> 00:07:25.520 Neither of our two governing equations contains x or z at all any more. 00:07:26.131 --> 00:07:29.867 Put the two equations through that transform and what is left in each one 00:07:29.950 --> 00:07:34.978 is an ordinary differential equation in y alone, with time as a parameter. 00:07:35.593 --> 00:07:39.715 Fourth order for the velocity, second order for the vorticity. 00:07:40.419 --> 00:07:43.612 And here is the payoff the whole method is built on. 00:07:44.064 --> 00:07:47.012 Each pair of wavenumbers gets its own pair of equations, 00:07:47.095 --> 00:07:50.892 and those equations know nothing whatsoever about any other pair. 00:07:51.205 --> 00:07:53.921 The three-dimensional problem has come apart into a lattice 00:07:53.921 --> 00:07:55.954 of one-dimensional problems. 00:07:56.750 --> 00:07:58.643 Walk across that lattice. 00:07:59.095 --> 00:08:00.547 Here is one mode. 00:08:00.744 --> 00:08:02.730 Here is another, further out. 00:08:02.927 --> 00:08:04.530 And here is a third. 00:08:04.842 --> 00:08:08.616 Three completely independent boundary value problems, and the only thing 00:08:08.616 --> 00:08:12.842 that differs between them is the single number k squared sitting in the operator. 00:08:13.546 --> 00:08:15.799 Where have u and w gone? 00:08:16.413 --> 00:08:17.447 Nowhere. 00:08:17.644 --> 00:08:19.328 They have become algebra. 00:08:19.780 --> 00:08:21.891 Continuity, transformed, 00:08:21.974 --> 00:08:26.225 is one linear relation between the three transformed components. 00:08:27.021 --> 00:08:29.817 And the definition of the wall-normal vorticity, 00:08:29.900 --> 00:08:34.290 transformed, is a second linear relation between the same three. 00:08:34.486 --> 00:08:38.084 In both of them the y derivative of v hat is already known, 00:08:38.167 --> 00:08:41.163 because v is what we have just marched. 00:08:41.646 --> 00:08:46.825 So set them side by side as a two by two system for u hat and w hat. 00:08:47.137 --> 00:08:51.782 Multiply down one diagonal, multiply down the other, and subtract. 00:08:52.095 --> 00:08:57.178 The determinant is k x squared plus k z squared, which is k squared: 00:08:57.261 --> 00:09:02.092 nonzero for every mode but one, and we will come back to that exception. 00:09:02.401 --> 00:09:05.815 Solve it and the horizontal velocities fall out. 00:09:06.012 --> 00:09:12.117 u hat is one combination of the slope of v hat and the vorticity, w hat is another, 00:09:12.200 --> 00:09:14.952 and both are divided by k squared. 00:09:15.865 --> 00:09:19.941 No linear solve, no iteration, no elliptic problem: 00:09:20.045 --> 00:09:24.074 two multiplications and an addition, per mode, per step. 00:09:24.178 --> 00:09:27.290 And continuity is not being enforced here at all. 00:09:27.487 --> 00:09:28.485 It has been used, 00:09:28.485 --> 00:09:33.223 so the reconstructed field satisfies it identically, to machine precision. 00:09:34.136 --> 00:09:37.526 One thing is still missing, and it is not a small thing. 00:09:37.723 --> 00:09:40.415 This reconstruction divides by k squared, 00:09:40.498 --> 00:09:44.655 and there is exactly one mode in the lattice where k squared is zero. 00:09:44.851 --> 00:09:48.149 That mode, and the boundary conditions, come next. 00:09:49.259 --> 00:09:54.147 A fourth-order equation wants four boundary conditions, two at each wall. 00:09:54.251 --> 00:09:56.806 A second-order equation wants two. 00:09:56.910 --> 00:10:01.458 So we need six, and the walls appear to offer only one thing: 00:10:01.542 --> 00:10:03.470 the velocity vanishes. 00:10:04.267 --> 00:10:06.125 Start with what is given. 00:10:06.322 --> 00:10:11.195 At either wall all three components are zero, for every x and every z, 00:10:11.279 --> 00:10:15.854 so mode by mode every one of the three amplitudes is zero as well. 00:10:16.651 --> 00:10:18.808 That is v hat equal to zero 00:10:18.892 --> 00:10:23.165 at both walls: two conditions for the fourth-order equation. 00:10:23.269 --> 00:10:28.285 For the other two, put the wall values into the transformed continuity relation. 00:10:28.481 --> 00:10:31.195 Since u hat and w hat vanish there, 00:10:31.279 --> 00:10:35.262 the y derivative of v hat has to vanish there too. 00:10:36.477 --> 00:10:39.032 So the picture at each wall is this: 00:10:39.136 --> 00:10:42.387 v hat pinned to zero, and pinned flat. 00:10:42.583 --> 00:10:46.938 Four conditions, and not one of them was invented for numerical convenience. 00:10:47.734 --> 00:10:50.127 The vorticity is easier still. 00:10:50.323 --> 00:10:52.669 Wall-normal vorticity is built from u 00:10:52.669 --> 00:10:58.463 and w alone, and both of those are zero at the wall, so omega y hat vanishes there. 00:10:58.659 --> 00:11:01.667 Two conditions for a second-order equation. 00:11:02.580 --> 00:11:04.226 Let me count it once more, 00:11:04.310 --> 00:11:07.956 because this is the part that makes the formulation feel inevitable. 00:11:08.152 --> 00:11:12.646 Each unknown, its order in y, and what the walls hand us for it. 00:11:13.443 --> 00:11:17.553 The wall-normal velocity obeys a fourth-order equation, and the walls 00:11:17.553 --> 00:11:23.463 give exactly four conditions: the value, and the slope, at each of the two walls. 00:11:24.515 --> 00:11:28.065 The wall-normal vorticity obeys a second-order equation, 00:11:28.149 --> 00:11:30.600 and the walls give exactly two. 00:11:30.796 --> 00:11:33.816 Nothing is left over, and nothing has to be guessed. 00:11:34.128 --> 00:11:38.889 Compare that with a boundary condition for the pressure, which does not exist. 00:11:39.686 --> 00:11:41.289 Now the exception. 00:11:41.903 --> 00:11:45.848 The reconstruction of the horizontal velocities divided by k squared, 00:11:45.932 --> 00:11:49.308 and there is exactly one mode where k squared is zero: 00:11:49.392 --> 00:11:53.073 no variation in x and no variation in z. 00:11:53.177 --> 00:11:57.295 That mode is the mean flow, averaged over wall-parallel planes, 00:11:57.379 --> 00:12:00.805 and everything we have built so far says nothing about it. 00:12:01.602 --> 00:12:03.727 One of its components is free. 00:12:03.831 --> 00:12:08.669 Continuity for that mode says the y derivative of v hat is zero, 00:12:08.753 --> 00:12:14.838 and v hat is zero at the wall, so v hat is identically zero right across the gap. 00:12:14.942 --> 00:12:19.749 There is no mean flow through the walls, which is exactly what we should expect. 00:12:20.546 --> 00:12:23.852 The two horizontal means are not determined algebraically, 00:12:23.936 --> 00:12:26.003 so they are marched directly. 00:12:26.316 --> 00:12:28.278 Average the x momentum equation 00:12:28.278 --> 00:12:32.643 over a wall-parallel plane and you get a one-dimensional diffusion equation 00:12:32.643 --> 00:12:38.136 for u bar, forced by the mean nonlinear term and by the mean pressure gradient. 00:12:39.048 --> 00:12:43.005 And there is the one piece of the pressure that survives the whole construction: 00:12:43.089 --> 00:12:48.349 a single number, the mean streamwise gradient, which is what drives the flow. 00:12:48.546 --> 00:12:52.737 Either you fix it and let the flow rate settle, or you fix the flow rate 00:12:52.737 --> 00:12:54.642 and adjust it every step. 00:12:55.345 --> 00:12:59.804 The spanwise mean gets the same treatment with no imposed gradient. 00:13:00.001 --> 00:13:04.062 Both are one-dimensional diffusion equations with zero at each wall, 00:13:04.146 --> 00:13:07.141 and the machinery of the next part solves them without noticing 00:13:07.141 --> 00:13:09.406 that anything special has happened. 00:13:10.543 --> 00:13:14.046 The wall-normal direction is neither periodic nor homogeneous, 00:13:14.130 --> 00:13:16.732 so Fourier is the wrong basis there. 00:13:16.928 --> 00:13:19.526 Expand in Chebyshev polynomials instead, 00:13:19.610 --> 00:13:23.953 on the interval from minus one to plus one, which is exactly the gap. 00:13:24.750 --> 00:13:26.759 The first few look like this. 00:13:26.863 --> 00:13:32.889 T one is just y. T four already has four zeros inside the gap. 00:13:32.993 --> 00:13:36.604 T eight has eight, and notice where they crowd: 00:13:36.708 --> 00:13:39.658 the oscillations bunch up towards the two ends. 00:13:40.454 --> 00:13:46.396 So the expansion is a finite sum, N plus one coefficients for each wavenumber pair, 00:13:46.480 --> 00:13:49.882 and those coefficients are what the code actually stores. 00:13:50.679 --> 00:13:54.859 The collocation points come with the basis rather than being chosen. 00:13:55.172 --> 00:13:57.735 They are cosines of equally spaced angles, 00:13:57.819 --> 00:14:01.361 and here they are across the gap for N equal to eight. 00:14:01.902 --> 00:14:03.386 Look at the spacing. 00:14:03.470 --> 00:14:07.023 In the middle of the channel the points are as coarse as they ever get. 00:14:07.564 --> 00:14:09.837 Against the wall they are packed tight, 00:14:09.921 --> 00:14:13.080 with a spacing that shrinks like one over N squared. 00:14:13.183 --> 00:14:18.350 That is precisely the grid a wall-bounded flow wants, because the whole difficulty 00:14:18.350 --> 00:14:21.903 of this problem is a thin viscous layer against each wall. 00:14:22.607 --> 00:14:28.006 Now, what does the operator D squared minus k squared look like in this basis? 00:14:28.110 --> 00:14:32.821 Written for the coefficients, either as a tau formulation or as a Galerkin one, 00:14:32.905 --> 00:14:34.983 the second derivative connects a coefficient 00:14:34.983 --> 00:14:37.085 to only a few of its neighbours. 00:14:37.537 --> 00:14:39.268 The matrix is banded. 00:14:40.064 --> 00:14:43.432 For the second-order operator that means three diagonals. 00:14:43.535 --> 00:14:46.122 For the fourth-order operator, five. 00:14:46.206 --> 00:14:49.152 Either way it is a fixed number of entries per row, 00:14:49.236 --> 00:14:54.949 whatever N is, so one solve costs order N operations rather than order N cubed. 00:14:55.745 --> 00:14:59.972 Two further economies, and they are the ones that matter in practice. 00:15:00.169 --> 00:15:05.518 The matrix depends on the wavenumbers only through k squared, and on the time step, 00:15:05.602 --> 00:15:08.900 and neither of those changes as the simulation runs. 00:15:09.097 --> 00:15:10.499 So it is factored once, 00:15:10.583 --> 00:15:14.984 before the first step, and every step afterwards is a back substitution. 00:15:15.781 --> 00:15:18.104 And one more, which is free. 00:15:18.417 --> 00:15:21.525 The operator does not mix even polynomials with odd ones, 00:15:21.609 --> 00:15:25.894 so each system splits into two half-size systems, one per parity. 00:15:26.091 --> 00:15:29.667 All of that, for every wavenumber pair independently. 00:15:30.787 --> 00:15:33.594 Here is one time step for the vorticity equation, 00:15:33.678 --> 00:15:36.152 written the way it is usually implemented. 00:15:36.348 --> 00:15:38.482 The viscous term is treated implicitly, 00:15:38.566 --> 00:15:42.897 by Crank-Nicolson: half of it at the old level, half at the new. 00:15:43.810 --> 00:15:49.105 The nonlinear source is treated explicitly, by a two-step Adams-Bashforth: 00:15:49.208 --> 00:15:53.342 three halves of the current value, minus a half of the previous one. 00:15:53.539 --> 00:15:57.278 No solve is involved, because both are already known. 00:15:57.982 --> 00:16:00.525 And look at what the left hand side is. 00:16:00.629 --> 00:16:04.027 It is the banded, prefactored operator from a moment ago, 00:16:04.111 --> 00:16:07.917 one per wavenumber pair, unchanged from step to step. 00:16:08.001 --> 00:16:13.168 The whole advance is a back substitution per mode, plus the work of forming h. 00:16:13.872 --> 00:16:18.679 Now the stability, which is the real reason for treating viscosity implicitly. 00:16:18.875 --> 00:16:23.122 The time step is no longer limited by the wall-normal spacing at all. 00:16:23.206 --> 00:16:26.816 That spacing goes like one over N squared, so an explicit 00:16:26.816 --> 00:16:30.996 viscous treatment would need a step going like one over N to the fourth. 00:16:31.100 --> 00:16:34.178 On a fine grid that is simply unusable. 00:16:35.090 --> 00:16:38.063 The velocity equation needs one more idea. 00:16:38.260 --> 00:16:39.966 It is fourth order, so rather 00:16:39.966 --> 00:16:44.414 than building a fourth-order operator we split it into two second-order problems. 00:16:44.610 --> 00:16:48.024 First solve for phi, the Laplacian of v hat. 00:16:48.337 --> 00:16:52.378 Then solve for v hat itself, with phi as the source. 00:16:53.175 --> 00:16:55.916 There is a catch here, and it is worth naming. 00:16:56.112 --> 00:16:59.511 All four boundary conditions are conditions on v hat: 00:16:59.595 --> 00:17:01.941 two values and two slopes. 00:17:02.254 --> 00:17:05.435 Neither of the two second-order problems has any natural condition 00:17:05.435 --> 00:17:07.073 on phi at the walls. 00:17:07.776 --> 00:17:10.595 The standard remedy is a Green's function argument. 00:17:10.679 --> 00:17:14.821 Solve the pair once with the real forcing and zero conditions on phi, 00:17:14.905 --> 00:17:19.131 then twice more with unit conditions and no forcing, and take the combination 00:17:19.131 --> 00:17:23.009 of the three that makes the slope of v hat vanish at both walls. 00:17:23.206 --> 00:17:27.920 That is a two by two solve per mode on quantities computed before the run. 00:17:28.717 --> 00:17:33.103 That leaves the source terms h, which are the only nonlinear work in the method, 00:17:33.187 --> 00:17:35.951 and they are evaluated pseudospectrally. 00:17:36.403 --> 00:17:39.306 Transform the modes back to a physical grid. 00:17:39.410 --> 00:17:43.335 Form the products there, where a product is just a multiplication. 00:17:43.531 --> 00:17:46.249 Then transform the result back to modes. 00:17:47.162 --> 00:17:50.077 Why not form the products mode by mode instead? 00:17:50.389 --> 00:17:55.405 Because a product of two Fourier series is a convolution, which costs order N squared 00:17:55.405 --> 00:18:01.234 per direction, while two transforms and a multiplication cost order N log N. 00:18:01.546 --> 00:18:06.110 That single choice is what makes spectral direct simulation affordable at all. 00:18:06.651 --> 00:18:08.567 The price is aliasing. 00:18:08.880 --> 00:18:14.116 A product of two modes inside the retained band generates content outside it, and on 00:18:14.116 --> 00:18:19.063 a finite grid that content folds back onto the modes you are keeping, as error. 00:18:19.859 --> 00:18:23.966 So in both wall-parallel directions the products are formed on an enlarged grid 00:18:24.050 --> 00:18:28.277 and everything above the retained band is thrown away rather than folded back. 00:18:28.474 --> 00:18:30.317 That is the three-halves rule, 00:18:30.401 --> 00:18:34.164 and a phase shift scheme does the same job by a different route. 00:18:34.705 --> 00:18:37.082 So let me put the efficiency in one place. 00:18:37.166 --> 00:18:40.275 First: there is no elliptic pressure solve, at any step, 00:18:40.359 --> 00:18:42.217 because there is no pressure. 00:18:42.530 --> 00:18:46.812 Second: the reconstructed velocity is divergence free by construction, 00:18:46.896 --> 00:18:50.229 so nothing has to be projected or corrected afterwards. 00:18:51.025 --> 00:18:55.225 Third: the Fourier transform in the two homogeneous directions 00:18:55.309 --> 00:18:59.451 uncouples the three-dimensional problem into an independent, banded, 00:18:59.535 --> 00:19:03.402 already-factored boundary value problem for every wavenumber pair. 00:19:03.599 --> 00:19:06.118 Fourth: every boundary condition the method 00:19:06.118 --> 00:19:10.879 needs is one the walls actually give, so there is nothing invented near the wall. 00:19:11.676 --> 00:19:15.833 What is left, per step, is dominated by the transforms: 00:19:16.285 --> 00:19:19.591 order N log N in the wall-parallel directions, 00:19:19.675 --> 00:19:22.265 times the number of points across the gap. 00:19:22.369 --> 00:19:26.561 The solves themselves are linear in the number of Chebyshev modes. 00:19:27.264 --> 00:19:28.925 That is the whole of it. 00:19:29.122 --> 00:19:31.259 Two curls removed the pressure. 00:19:31.456 --> 00:19:35.462 Fourier removed two of the three space dimensions from every solve. 00:19:35.566 --> 00:19:38.466 Chebyshev put the points where the walls need them. 00:19:38.550 --> 00:19:40.152 And no-slip turned out to supply 00:19:40.152 --> 00:19:44.077 exactly the six boundary conditions the two equations were waiting for.