## The Kuramoto–Sivashinsky Equation (Part 7)

I have a lot of catching up to do. I want to share a bunch of work by Steve Huntsman. I’ll start with some older material. A bit of this may be ‘outdated’ by his later work, but I figure it’s all worth recording.

One goal here is to define ‘stripes’ for the Kuramoto–Sivashinky equation in a way that lets us count them, their births, and their mergers, and so on. We need a good definition to test the conjectures I made in Part 1.

While I originally formulated my conjectures for the ‘integral form’ of the
Kuramoto–Sivashinky equation:

$h_t + h_{xx} + h_{xxxx} + \frac{1}{2} (h_x)^2 = 0$

Steve has mostly been working with the derivative form:

$u_t + u_{xx} + u_{xxxx} + u u_x = 0$

so you can assume that unless I say otherwise. He’s using periodic boundary conditions such that

$u(t,x) = u(t,x+L)$

for some length $L.$ The length depends on the particular experiment he’s doing.

First, a plot of stripes. It looks like $L = 100$ here:

Births and deaths are shown as green and red dots, respectively. But to see them, you may need to click on the picture to enlarge it!

According to my conjecture there should be no red dots. The red dots at the top and the bottom of the image don’t count: they mostly arise because this program doesn’t take the periodic boundary conditions into account. There are two other red dots, which are worth thinking about.

Nice! But how are stripes being defined here? He describes how:

The stripe definition is mostly pretty simple and not image processy at all, and the trick to improve it is limited to removing little blobs and is easily explained.

Let $u(t,x)$ be the solution to the KSE. Then let

$v(t,x) := u(t,x)- u(t,x+a)$

where $a$ is the average integer offset (maybe I’m missing a minus sign a la $-a$) that maximizes the cross-correlation between $u(t,x)$ and $-u(t,x+a)$. Now anywhere $v$ exceeds its median is part of a stripe.

The image processing trick is that I delete little stripes (and I use what image processors would call 4-connectivity to define simply connected regions—this is the conservative idea that a pixel should have a neighbor to the north, south, east, or west to be connected to that neighbor, instead of the aggressive 8-connectivity that allows NE, NW, SE, SW too) whose area is less than 1000 grid points. So it uses lots of image processing machinery to actually do its job, but the definition is simple and easily explained mathematically.

An obvious fix that removes the two nontrivial deaths in the picture I sent is to require a death to be sufficiently far away from another stripe: here I am guessing that the characteristic radius of a stripe will work just fine.

### 6 Responses to The Kuramoto–Sivashinsky Equation (Part 7)

1. Steve Huntsman says:

I think the biggest shift in my thinking of late has been to describe the lifetimes of stripes. Essentially, they are either “evanescent” and die shortly after birth, or they are (practically? numerically? I haven’t yet focused on statistics) immortal. I have ditched the idea of defining/image processing-away the evanescent stripes in favor of quantifying them in this way, which is both simpler to define and more granular.

My major sticking point at present, and the cause of two late nights in a row, has been to find a computationally efficient way to determine the lifetimes of splits (versus stripes per se). I haven’t been able to exploit the “adjunction” births merges / deaths splits in this particular regard, though it works fine in identifying the loci of events.

Anyway if one ignores (or perhaps it might be better to say properly accounts for) the evanescent stripes I claim there is preliminary support for John’s quantitative conjectures.

• John Baez says:

Interesting! What do you mean by the lifetime of a split?

• Steve Huntsman says:

Well when a stripe splits, typically one branch just persists for a few temporal grid steps (or horizontal pixels if you prefer). The lifetime of a split is the time between the split itself and the first death in a resulting branch. Note that we can’t really guarantee the death “of” a resulting branch versus “in”, because you could have a series of evanescent splits. I’m not sure offhand if my code actually does “in” or “of”–I’ll have to circle back to that, but it’s tricky enough as is that I wouldn’t try to adapt it further.

Rather than try to tweak the definition of a stripe, I think it’s more sensible to demonstrate that any phenomenological unpleasantness is fleeting, whether it’s for deaths (birth was typically just a moment ago) or splits (death is typically just a moment later).

I’m sure you’ve noted the “adjunction/dagger” between a (binary) stripe diagram and the negation of its horizontal flip–this restricts to births/deaths/merges/splits. My code tries to exploit this as much as possible, but a split lifetime seems to require a different treatment than a birth lifetime for topological reasons.

2. domenico says:

I try to obtain the Lagrangian and Hamiltonian of Kurasamoto-Sivashinsky equation.
An extremal Lagrangian for the equation is
$L = \left(h_t+h_{xx}+h_{xxxx}+\frac{1}{2}h^2_x\right)^2$
because of only along the trajectory the Lagrangian has a global minimum.
The Euler-Lagrange equation is:
$0=-\frac{d}{dt}\left(h_t+h_{xx}+h_{xxxx}+\frac{1}{2}h^2_x\right)$
that is the derivative of the differential Kurasamoto-Sivashinsky equation, so that it contain the solution of the equation.
The Hamiltonian of the Kurasamoto-Sivashinsky equation is simply (I don’t use the Legendre trasformation of the Lagrangian but the equations of motion):
$H = -p\left(h_{xx}+h_{xxxx}+\frac{1}{2}h^2_x\right)+f(h)$
where f(h) is an arbitrary function, the Hamilton’s equation are:
$h_t=-\left(h_{xx}+h_{xxxx}+\frac{1}{2}h^2_x\right)$
$p_t = -\frac{\partial f}{\partial h}$
so that it is possible to write the Schrodinger equation
$i \hbar \partial_t \Psi = -i \hbar \left(h_{xx}+h_{xxxx}+\frac{1}{2}h^2_x\right) \partial_h \Psi +f(h) \Psi$
where I have a problem with the arbitrariness of the f(h).
The Hamiltonian dynamics is arbitray along the p axis, in the classical dynamics, so that all the integral invariant are preserved in the motion.
I try to obtain Lagrangian and Hamiltonian to obtain the conserved quantity, but the problem is that in the Noether‘s theorem the continuous simmetry is on x, and the variable here is h.

• John Baez says:

Hmm, this approach to getting a Lagrangian is new to me: it seems you’re taking the differential equation, writing it in the form

$X = 0$

and then using $X^2$ as the Lagrangian. But when you work out the Euler–Lagrange equation from this Lagrangian, it seems you don’t get $X = 0$ back.

What happens if you do this with the heat equation?

• domenico says:

The heat equation is:
$u_t = \Delta u$
the Lagrangian is
$L = (u_t-\Delta u)^2$
then if u(t,x) and x does not depend on t
$0 = \frac{\partial L}{\partial u}-\frac{d}{dt} \frac{\partial L}{\partial u_t}=-\frac{d}{dt} (u_t-\Delta u)$
if there are equation with u, then the action has ever a extremal along the trajectory, because of the lagrangian is ever null along the trajectory, and it is not null for different trajectories: if there exist a term $L_u$, then the Euler Lagrange equation has a solution that is the same differental equation:
$0 = \frac{\partial F^2}{\partial u}-\frac{d}{dt} \frac{\partial F^2}{\partial u_t}=2 F \frac{\partial F}{\partial u}-2 \frac{d}{dt} \left(F\frac{\partial F}{\partial u_t} \right)$
where F=0 is a solution. I am sure that this Lagrangian is right, and I am not perfectly sure about the Euler-Lagrange equation, because of it is possible to have Lagrangian of higher order $L(u,u_t,u_x,u_{xx})$ with these generalized variables, with Euler-Lagrange of higher order.

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