As you know, I find it beautiful how pervasive partial differential equations are throughout the scientific enterprise.
And this is certainly the theme of the remainder of the course, exploring finite differences and other methods for solving pdes.
Let me begin by passing around, actually for the second time, a pair of very nice books on this subject.
So a few lectures ago I passed around a stack of books, two of which were also about PD ease.
And I want to remind you of those two.
One is by Aria Isserlis called a first course in the numerical analysis of differential equations, and the other is by Randy Leveque.
Finite Difference Methods for Ordinary and Partial Differential Equations.
These are both fantastic books by leaders in the field, which will tell you, of course, much more than we cover in these lectures.
Now, whenever you talk about a numerical solution of pdes questions of stability and instability, numerical stability and instability come up.
And that's what drives a lot of what we do. And it's pretty interesting.
So the next section will call numerical instability.
And today, you'll see how in a very simple way, it controls the kind of algorithms that work for Pdes.
So let's see. Numerical instability.
So to remind you, last lecture, we looked at the heat equation, which is you t equals you x a u x x.
So the heat equation in one d would be u t equals u x x.
And we disparities that and showed how. Sometimes you need a very small time step.
So now let's consider the fourth order version of that u t equals.
You x x x x. Except the way I've written, that would be an exponentially ill posed problem.
You need a minus sign. The way it works with diffusion is four orders two, six, ten, 14, and so on.
A plus sign is diffusive and four orders for eight, 12 and so on.
A minus sign is diffusive. As you can see by plugging in a 48 mode.
So this is the physically right approach to a fourth order diffusion problem.
Now, let's suppose we want to discard ties that by a finite difference approximation, we do the obvious thing which will work.
And well, we note that you x x x x is the same as u x x x x.
So we can take the second difference of the second difference to get a pretty good approximation to that.
So let's approximate you x x x x by.
Something divided by eight squared and that's something we'll be three terms, one term minus twice another term plus a third term.
And each of those will be a finite difference. So this will be the J plus two minus to the J plus one plus the J might be j over h squared.
And this will be the j plus one minus 2vj plus the j minus one over h squared.
And this will be the J minus to the j minus one plus v j minus two over h square.
So we've taken the obvious. Second difference of the obvious.
Second difference to get the obvious fourth difference.
And that's a perfectly good way to describe the fourth derivative if you want to write it in matrix notation.
You could think of that as if we're going to apply that to compute something.
We're going to end up with a new time step, a vector of values being equal to a matrix times, the current time step.
So this would be. The Matrix representation.
Of an explicit as opposed to implicit.
Finite difference approximation to our equation.
So we've just arrived. A perfectly good method that will work.
And now let's run it. You can see the code is called M 39.
Fourth order diffusion. Just glancing at the code briefly.
I use the usual trick in these demo codes that I actually do use a matrix which would be crazy if it were a dense matrix,
but I use a sparse matrix, so it's not crazy. We set up a grid.
The space size H is 1/40 and then you can see that we go through this time stepping with that time step of K and the code
is set up so that the user inputs the time step so we can see the behaviour for different choices of the time step.
And of course all the work is done just in the wire loop.
You equals eight times you at every time step. So let's run that.
I'll say m 39, fourth order diffusion.
There's the initial distribution.
And then it asks me what time step I want. So I'm going to put in a very small time step.
I'll put in one E minus eight.
And let me make that visible. Okay.
So we put in a timestamp of ten to the minus eight. And here's what happens.
The usual principle of diffusion that it's interesting a little bit and then it gets boring pretty fast.
So there's fourth order diffusion.
You can see if you're interested in the physics of these things that it is a little different from the second order case.
The magnet tenacity properties are rather different. But anyway, there we have a perfectly reasonable solution of sorts.
Let's increase the time step. So I'll change that to.
Four times ten to the minus eight. Same thing.
Time step four times bigger. Now I'll change it to 4.8, eight times ten to the minus eight.
4.88 D minus eight. Little wiggly.
This is actually just barely stable, maybe. But of course, you can see that there are oscillations that aren't going away very quickly.
So that's very marginally stable. And of course, I'm now going to do it with 4.90 times ten to the minus eight, and the Wiggles will grow.
So that's about as marginal an instability as you could ask for in a linear problem.
And instability will grow at a good, clean exponential rate.
There you have a good clean exponential growth.
And of course, if I take a bigger time step, it'll only get worse if I took ten to the minus seven, for example.
Then it blows up. So we see there, again, the kind of instability that one sees in all sorts of schemes.
What's new is that because it's a fourth order problem. The critical time step is even smaller than before.
It's very, very small. Ten to the minus eight. And what's going on there is that this is a stiff problem.
The fourth derivative couples together, things at all frequencies.
And there's the high frequencies cannot be forgotten.
They're moving so quickly that it's causing the problems we associate with stiffness.
So something has to be done in order to improve that before we do anything.
Let's analyse it and understand what's going on. So let's do the standard for a trick that people use to analyse instability.
So this idea basically was introduced by von Neumann in the 1940s.
Computers were invented. Los Alamos was discovering all sorts of things, sort of during and then after the war.
And von Neumann was a key figure.
He was really one of the great geniuses of the 20th century, and he knew everything about pure and applied mathematics.
It's amazing how he did the whole range, from pure to applied.
He knew Fourier analysis very well, so when he saw instabilities like that, he of course knew that the thing to do was for analysis.
Here's how we do it. We're going to plug in a Fourier mode and see what happens to it as time passes.
So. Let's do our trial solution and we'll say, what if?
At one time? Steppe We have a pure forage mode.
So suppose that step end the solution looks like this.
The J at step n. I'll call it a sine wave, but I'll actually write down a complex exponential suppose at time at step and it's e to the i.
C. Times. SJ Which is another way of saying E to the.
I see. JH So we're doing very standard for analysis in space here.
C is the wave number. Which tells you it's the spatial frequency.
And we want to see how different wave numbers evolve as we take steps with the finite, different scheme.
So we plug this into the finite difference scheme and we see what the and plus one looks like.
So at step end plus one.
Well, the miracle of Fourier analysis is that this sine wave will be multiplied by a constant.
What we'll find is that the end plus one will be simply equal to some number, depending on the wave number.
So we call it JFC Times the and.
And G of C is known as the amplification factor.
So stability analysis in this business is about studying amplification factors.
And I need a new red pen. Yeah.
Okay. Well, so let's figure out what that amplification factor is.
We just plug the sine wave into the finite difference formula and here's what we get g of T is equal to.
And I'll just write it down once here. Practised with this stuff.
It's amazingly quick. We're going to plug this into there.
And what we find is that our galaxy is equal to one minus two times step over the fourth power of the space step.
And then we get various terms from those various things.
The v j plus two term gives you an e to the 2ij2i see h the v j plus one terms give you a minus four e to the i c h the vee-jay term in the middle.
It gives you a plus six e to the zero.
The the j minus one term gives you a minus four e to the minus.
I see h and finally, the leftmost term gives you a plus one.
E to the minus two i. C.
H. I'm a little embarrassed that I never wrote down up above.
So let me quickly do that. I never simplified that formula, which would have given me.
The J plus two minus for the J plus one plus six the J minus for v j minus one plus v j minus two.
So all of these collapse into that. And it's when I plug the sine wave into there.
Here are these numbers. One minus four, six minus four, one.
So there's the amplification factor as a function of the wave number C four for the fourth order disk realisation.
And again, of course we could well what we do is we look at all possible wave numbers and see how they behave.
Are they amplified, are they decayed? So let's simplify this a bit.
These things turn into one minus K over H to the fourth times, the fourth power of E to the i c h over two minus e to the minus i.
C h over to. That's just algebra.
And if you look at these things, you see a sine wave, right? It's an E to the eye, something minus E to the minus eye something.
So this in turn is one minus K over H to the fourth times.
The sign of c h over two squared to the fourth.
So we have the fourth power of a sine wave dependence on the wave number.
And of course, the fourth power is always a positive number. So we have one minus a positive number.
And the question for stability is, is that thing less than one in which case will get decay or is it bigger than one?
In which case we'll get instability and growth.
So that's the key question of von Neumann analysis is this.
Less than one in absolute value. For all wave numbers.
If he's good, if no bad. Well, that's pretty easy to figure out.
The sign as ranges over numbers between minus one and one.
So the fourth Power Rangers, our numbers between zero and one.
So this will range over numbers to the left of one.
On the real axis, there's one. As sea varies, that will always take values one or two, the left.
And how far to the left will it go? It will go to one minus K over H to the fourth.
So there is this interval of the real axis. That will be the amplification factor for different wave numbers.
We want to know, is that interval always less than one an absolute value?
So the question becomes is, is it the case that one minus K over to the fourth is bigger than minus one?
That is our stability condition based on von Neumann analysis.
So in other words gets to simplify that is to bigger than 16 K over to the fourth i.e. is K less than or equal to H to the fourth over eight.
That is our von Neumann stability condition. But.
16. Where did the 16 come from?
There should have been a two to the fourth. I forgot the two to the fourth, didn't I?
Thank you. So this is to high times the sign.
It's not just the sign. It's two high times the sign. Thank you. Two are well spotted.
So hopefully when you work that out. So of course I've done this very quickly.
But in outline, what you can see is a very simple procedure for analysing these irregular grids and linear problems.
Plug in for your mode to see if they can grow or not.
You reach a conclusion, and that's a pretty pessimistic conclusion.
It says that if you have the space step, you have to shrink the time step by a factor of 16.
So you're going to need very small time steps, which, of course, is exactly what we found with the experiment.
In fact. If H equals 1/40, which is what we had in the code.
Then what you find is that the stability restriction is less than 4.88, three times ten to the minus eight.
So 4.88 is stable. 4.90 is unstable.
So we have a stiff problem here. The high frequencies are giving us trouble in some sense.
And actually, next week, we're going to talk in a general way about how one addresses the stiffness of pdes.
For the moment, let's talk about the basic tool that we have for that which is implicit as opposed to explicit, different thing.
Just like he's. We want to use implicit ness to allow ourselves to be less constrained in the time step.
So implicit one the finite differences.
And the basic thing that you start by doing is exactly what we did with ODS.
You use a backward rather than a forward formula.
Now, I didn't draw the stencil, in fact, but I could have done that when we wrote down our finite difference formula.
Where did we write down our finite?
I guess we never quite wrote it down, but I hope you understood that we were coupling together five points at step RN and computing from them.
A point at step and plus one. So that's the stencil of this explicit formula.
Well, to get around the stability restraint,
we want to use an implicit formula that will change that picture to find a different thing in the nude in the future, coupled with the present.
That will be our new stencil. And we can immediately write down the formula.
Let's do that. So our implicit backward Euler style disorganisation of the fourth order problem will be v, j plus v and plus one equals v.
N. And then we have minus because it's minus the fourth derivative K over H to the fourth.
And then we have these five terms, the J plus two minus for the J plus one plus six, the J minus for the J, minus one plus the J minus two.
The explicit formula had nn, but the implicit formula will have N plus one and plus one and so on.
So that's the. Formula for the obvious implicit descriptors in the backward Euler style disorganisation.
To analyse it, we do the same thing.
We plug in a R and that's of a four way mode, a sine wave, and we see by what constant is it multiplied as you go from one step to the next.
And I won't spell that out, but you get the very same result, except that the sign terms are in the denominator rather than the numerator.
You find that the amplification factor. Is now equal to one over.
One plus 16 K over H to the fourth.
Times. The sign of c h over to.
To the fourth. Hope I've got that right. You've got me nervous now.
All right. What did we get before? Yes.
So although I neglected the 16, as you pointed out, this is we before we had that in the numerator and now we've got it in the denominator.
But the good news is this is less than one no matter what.
For all see, regardless of K.
In fact, the amplification factor, as you can see, are always one or bigger sorry, one, one or smaller.
And so nothing can ever blow up.
This will be a stable formula for any timestamp, and the phrase people use is unconditionally stable.
No conditions on the time stamp. So of course, that doesn't mean you can use any time stamp,
because if your time step is too big, it'll still be inaccurate, but at least it won't blow up.
And we know the sort of general theme of numerical computing with differential equations is that to get convergence,
you need accuracy and you need stability. Well, this means that the only thing we have to check is the accuracy to get a good solution.
So of course we have a code for that. If you look at M40, you can see it's just like M 39.
And the only change is that we've changed a times U in the main loop to be backslash U in the main loop.
Now, with that high level of abstraction, it's a change of about two characters.
Of course, B is marginally different, but not very different. The signs are just a little bit different.
So we've turned matrix multiplication into solving a system of equations.
And that's doable because in this one dimensional problem, it's a penta diagonal system of equation.
So this thing here. We have V and plus one times B equals VM.
So we're solving a system of equations to get the new value.
But it's a painted diagonal system of equations.
Five non-zero diagonals, all the others zero and MATLAB is of course clever enough to do a good algorithm for that.
It's o of n operations for an n by n point a diagonal matrix.
So this very simple sparsity pattern is easily taken advantage of to get a fast solution.
So we run the code. And of course, I hope you know what we'll see.
Nothing will ever blow up. So this one is called M 40 implicit.
And again, I can choose the time step. So if I say ten to the minus eight, the solution will be almost identical to what we saw before.
And the speed the speed of computation isn't much different. It's not that hard to solve the system of equations.
You can sense as that evolves that it's an absurdly small time step.
It doesn't make sense to be doing all this work to get from one time step to the next.
If we increase that, we can go as far as we want.
So, you know, pick something ten to the minus five. So now we have a thousand times bigger time step.
Now you can see how the thing really evolves for a longer time, which we effectively couldn't see at all before.
So in terms of stiffness, remember the general definition of stiffness?
Is that a problem is if it contains time scales much shorter than the ones you are tracking.
So is this problem stiff? Well, to be precise, at the very beginning, it's not when you begin the calculation.
Because that is not a smooth curve. For the first few milliseconds, there's something going on, on a very small timescale step.
So for the first few billionth of a time unit, you would say that this is not a stiff PD.
But then after that, things are smooth. And so the equation has these fast time scales in it that are irrelevant to the thing you tracking.
So it quickly becomes a stiff problem. So let's just do that again.
If I say ten to the minus five again. You can see very quickly, it's gotten to the point.
Where the fast timescales have nothing to do with the evolution of the curve.
And of course, it's unconditionally stable. So I could take point one.
Yeah, it works. So we have this just absolutely huge difference between explicit and implicit.
Now everything becomes much more fun when you have nonlinearity there.
And it turns out the same ideas are at the heart of what is a practical way to cope with nonlinearity.
And we're going to talk about that a bit now with an example and then more next week.
So I want to tell you about one of my favourite PD e's, the motto Sufficiency Equation.
This is a prototype of a non-linear equation with stiff behaviour and the equation is u t equals minus u x x minus you.
X x x x. Minus you squared over two x.
So of course it's a one dimensional nonlinear PD and there's a page from the coffee table book there, which I hope you'll read when you come to it.
To remind you, the coffee table book has about 40 of these nicely finished pages and you can find them online at my website.
Just click on books. So this equation has three terms on the right and everyone has a different physics.
And the interplay of those three bits of physics is what makes it so beautiful, just to say in words what those three things are.
This is negative linear diffusion because of the minus sign.
So if you had just that equation, you two equals minus x x.
That's called the backward heat equation. It's completely opposed.
Things just blow up. And the higher the wave numbers, the faster they blow up.
This one. Has positive diffuse.
And now let's ignoring the nonlinear term. Let's think about what would happen to a Fourier mode plugged into that equation.
Well, this act to increase it exponentially and this acts to decrease it exponentially.
So let's try the trial solution for the nonlinear part.
All right. Sorry for the linear part of the equation. Suppose our trial solution is that you is equal to a four year mode, which I'll say is e to the.
I see x. And I'll give it some sigma time t.
I'm not putting an eye there because we're going to get not wavelike behaviour but explosively increasing or decreasing behaviour.
If you plug that into these two terms, well the minus u double x term we have Ike C squared,
so we get minus x squared and then with another minus sign.
So we get plus c squared. So we find that. You get?
You at s t. Well, you get that.
Sigma is related to see by this equation.
And this is a kind of a dispersion relation or a diffusion relation would be a better term,
but nobody ever speaks of diffusion relations at the dispersion relation.
For the behaviour of the linear part of the promoter of efficiency equation.
It tells you that small wave numbers where this term dominates have a positive sigma so they'll grow.
And large wave numbers where this term dominates have a negative sigma, so they'll decay.
And in fact, if I plot this as a function of C, you'll get this kind of behaviour.
When C equals one, you get a crossover.
Between a positive sigma and a negative sigma. So this is the regime of negative diffusion and this is the regime of positive diffusion.
The high wave numbers diffused. So if we just had that linear equation without the nonlinear term, it would be opposed.
So solutions, technically speaking, wouldn't really exist.
But anyway, the behaviour in question would be that the low wave numbers would simply explode and the high wave numbers would decay.
So you would see bigger and bigger, fairly smooth waves.
However. This term has the beautiful effect of moving energy from the low wave numbers to the high wave numbers.
So this is the nonlinear term.
What it does is it takes these relatively smooth components and stiffens them up, like the Burger's equation, which we mentioned earlier.
And I'm not going to try to write that down mathematically, but in effect, it takes low, smooth energy and turns it into less smooth energy.
So it's moving energy from this part of the picture to that part of the picture.
And because of that, you end up with a well posed PD with wonderfully complicated behaviour.
The you get the structures that are fairly smooth because this is where the energy is amplified.
But those structures don't blow up because as they get bigger, energy is siphoned off into the high wave numbers where it decays.
So it's a kind of an engine moving energy around, growing it in one region, decaying it in another.
Now to solve that numerically. This is an equation you don't particularly care about, although it's pretty,
but it really is a prototype of a lot of computational science with pdes to solve it numerically.
You need to treat the linear terms implicitly and the nonlinear terms explicitly.
You don't need to, but that's the best way to do it.
And that's almost the central, most important practical fact about how people solve pdes like this.
And let me summarise why it's important to do the linear terms implicitly and the nonlinear terms explicitly.
So the good strategy for this and many other problems.
Is that somehow or other? You want to do the linear terms implicitly and the nonlinear terms explicitly.
And why is that? Well.
Because it's the linear terms that are causing the stability constraint, this very high frequency and this fairly high frequency.
Those are the ones that would force you to have an absurdly small timestamp if you used an explicit formula.
So the reason the linear terms should be implicit is to avoid a very tight stability restriction.
A very tight limit on the timestamp. Why are the nonlinear terms explicit?
Because whenever you have an implicit formula, you have to solve some equations.
If they're linear equations, that's just linear algebra. But if they're nonlinear equations, that's more than linear algebra.
You have to do a Newton iteration or something crazy. That can be hard.
So explicit for the nonlinear terms to avoid having to solve a nonlinear problem at each step.
Now, the reason this strategy is so fruitful is that for all sorts of problems, the linear terms are the stiff ones.
They're the ones with the high derivatives, with the fast timescales.
The nonlinear terms are operate on a slower timescale.
In principle, it could be the other way around. If you had the nonlinear terms with the high derivatives, then you'd be in big trouble.
Then this strategy would get you now where you basically have to be implicit and do a lot of work at each time step.
But luckily a lot of problems are in this category where the stiff terms are linear and it's kind of amazing how easy it is,
at least for this one D problem to take advantage of that.
So what we're simply going to do is use a stencil.
That's implicit. For the fourth order term and the second order term.
But the nonlinear thing will not be at that level at all.
And we in feed in, if you like. The nonlinear term.
At times step n. So this, of course, as always, is the end plus one part of the stencil, and that's the time stamp and part of the stencil.
So you blend these together and let's just look at the code.
It's so simple what you need to do. So the code is called M 41 Courier Moto Z MUSZYNSKI Take a look at that.
You'll see that it's got the same flavour as all these last few codes I've been showing you.
It sets up a grid. It sets up some matrices. And then the key line in the code.
In the wire loop. Is this one that says you equals B backslash.
You minus a times you squared over two.
Now that is doing just what I mentioned.
The solution of a system of equations involving that point, a diagonal matrix is doing the second and fourth order linear turns.
And then the A matrix which is not being solved, it gets multiplying by something is of course the nonlinear term.
So given a vector you of values at time step in, we can evaluate that and then we can do a backslash.
And we've got our discrete approximation that the next time step and this works beautifully.
And let me show you the picture. The reason it's I like this equation so much is that the solutions are chaotic.
So I'll say M. 41.
Oops. Komodo said Wyszynski. So that's the initial condition.
And it asks me, what time do I want to go to? Well, let's make it a nice big time.
I'll say a thousand. So as I go out type return to see solution.
Okay, so it's evolving and the initial condition is doing complicated stuff.
And then you see this beautiful, seemingly random motion.
And notice. The interplay of randomness with tremendous structure.
So, of course, that looks random, but the structure is that it's clearly got a well-defined wavelength,
it has a preferred wavelength, and that will be determined by this curve.
The preferred wavelength is the one that's amplified the most. So energy gets gets increased at that fairly low wave number, fairly smooth wave.
But it can't get amplified too much because the nonlinear term then sends it off to the higher wave numbers where it gets diffused away.
And. People have proved that these solutions truly are chaotic.
There's a strange attractor. All the requisite features of chaos are there.
Including the exponential dependence on initial conditions.
So that picture is not in any sense. Well, it's not in the literal sense, correct?
At times 626, this is not the right curve because little errors along the way will have been
amplified exponentially according to the up and off exponent for this chaotic problem.
So none of these curves are literally correct.
After the first time, a few time units, we have that chaotic situation where the overall behaviour is correct,
but the precise detail at a precise time is not correct.
I can just stare at that stuff all day, but luckily we're only going to 1000.
Let's, however, run it a couple more times just for a brief time interval.
So suppose I run that again. If I guess, go to time ten, for example.
What you see is that the hump kind of increases and then it begins to wiggle.
So the increase is this linear effect of energy at low wave numbers being amplified.
But then the beginning of the Wiggles is this non-linear effect of it being transferred to higher wave numbers.
So wiggling that's that's the Burger's equation, the steepening up of a front.
Let's go a little further. Suppose I took it to term 20. So it grows and then wiggles.
Come in and then. You're beginning to sense complexity.
And of course, then it slowly gets chaotic. Now I want to show you.
The latest thing the club fund team has been doing, or rather our Adrian Martinelli has been doing, which is related to problems like this.
I'm going to show you how you can solve this problem. And Jeb, fun.
So let's. Well, the code there is m 41 seven, but I'm actually going to type it in so I can talk about it.
So the code we're speaking of is called spin.
And the reason it's called spin is that that stands for a stiff PD integrator.
It's not currently in the standard telephone, but you can get it by going to the branch feature spin up.
So the it tells you here what to do if you want to play with it.
So what you do is you declare what interval you're working on and the interval for this problem is 16 pi times minus one one.
So that's our domain. And then in fun style, you make a suitable operator.
We call it a spin up. This PDP operator defined on that domain.
And it has a linear part and a nonlinear part. So the linear part would be defined.
As an anonymous function in MATLAB, the usual fashion, and then in the standard seven fun style, we have these overloads of MATLAB commands.
So. The linear part is an anonymous function that has the negative of the second derivative and the negative of the fourth derivative.
And then the nonlinear part. Has to do that you squared effect.
So that will be an anonymous function. If anyone sees me type something wrong, please tell me.
What did they do in part? Oh yes.
And annoyingly, we have fights on the Champs team about this.
Currently, I'm losing. So you do have to get the case properly.
S dark non-linear part is equal to minus five times the.
Derivative of u squared and I. Yeah. Okay.
So in fact, there's now a thing called a spin up that we have now set up.
And to solve the problem, we, of course, needed the initial condition, which of course we give in the form of a telephone.
So I could say to have fun and then I could say cosine of x over 16.
Times one plus sign of the x over 16.
And then that's on the domain we specified.
And because it's periodic, I'll tell everyone that it's periodic by saying Trig, though, you don't have to do that.
And then we're going to evolve over some hour interval.
So that doesn't matter. Let's get hard. Line it. I'll say you.
Solution equals spin. So the spin up and then a time interval so we could say zero up to ten.
And then the initial condition you not so assuming I haven't mistyped anything more that will now call a certain algorithm,
which I'm not telling you about just now. And we will get much as what we got before.
Let's see, there's 1020 already on the screen. So let's change this to 20.
So we computed that with the finite differences, and now we're doing it with the spin code.
So please take an image in your brain of that curve and we'll see whether we get the same curve.
Okay. Type space when ready? I'm ready. Is that the same?
I hope so. It won't be the same at time. 500 because of the chaos.
But of course, for the initial times it should be the same. Now, this is actually a much more powerful system.
It's faster and more accurate. So although you can't really see just by looking at the picture, this is a much better solution than we had before.
I think it's stable enough that we could go to large times and nothing would go wrong.
Let's go to a slightly large time and see if that's true.
So next week I'm going to tell you about the algorithms that are used for this a little bit anyway.
And it also works in 2D and 3D.
So you can do various fun things with spin.
If you're interested in reaction to fusion equations in more dimensions, you can say things like spin.
Well, I could say Spin of Chaos, and I'd get the hardcore coated, comatose efficiency demo.
There it is. But then various other things are in there.
I could say KTV. Or I could say spin too.
And do something in multiple dimensions, say. I think that's the Ginsburg Landau equation.
Does that work? Did I make a mistake there?
Guess we have a two dimensional version of this evolving against Burton Lando equation that again has this linear,
non-linear structure and we're going to stop there for today.
So thanks.