# Periodicity of KdV equation in relation to zero-curvature equation

+ 4 like - 0 dislike
88 views

In most of the resources that I have read, integrable systems described by a PDE posses a zero-curvature equation $$\partial_t U - \partial_x V + [U,V] = 0$$ which gives rise to the monodromy matrix $$T = \mathcal{P} \exp \int\limits_0^{L} \mathrm{d} x \; U$$ Then the following quantities $$I_n(\lambda) = \mathrm{Tr} (T^n(t,\lambda))$$ are independent in time and so the system has an infinite amount of conserved charges, as required. However, to proof this, it is always (at least in all the notes and books I've read) assumed that all the fields are periodic in $x$ with period $L$ such that $$V(0,t,\lambda) = V(L,t,\lambda)$$ However, I really don't understand what is the justification for this assumption. For the KdV equation, for instance, it seems clear to me that $$u(x+L,t) \neq u(x,t) \forall x$$ which can be seen by simply looking at a plot of the (for instance 3-soliton solution) of the KdV equation

I might be misunderstanding something really simple because I've read many resources and they all make this assumption of periodicity without justifying it, but I hope someone can explain it to me.

This post imported from StackExchange MathOverflow at 2016-05-21 10:18 (UTC), posted by SE-user Hunter
asked May 18, 2016
retagged May 21, 2016
in order for the conserved quantities to be expressed as local functionals $F[u] = \int f(u,u_x,u_{xx},\ldots) dx$ you need to have boundary condition which allow to make sense of the integral. Periodicity is not always required, you can ask that $u(x) \to 0$ fast enough as $x\to \pm \infty$. Or you can define $u(x)$ on a circle ($x\in S^1$) so that integration makes sense and integrating by parts has no boundary term.

This post imported from StackExchange MathOverflow at 2016-05-21 10:18 (UTC), posted by SE-user issoroloap
@issoroloap I understand that we need some sort of boundary conditions, but for the proof that the quantities $I_n(\lambda)$ are conserved, they always impose periodic boundary conditions. For the KdV equation, these boundary conditions don't make sense to me, and so it seems that the proof doesn't make sense. I would be interested to see a proof for the boundary conditions where $u(x) \to 0$ fast enough as $x \to \pm \infty$. Or does the monodromy matrix then just becomes $T = \mathcal{P} \exp \int\limits_{-\infty}^{\infty} \mathrm{d} x \; U$?

This post imported from StackExchange MathOverflow at 2016-05-21 10:18 (UTC), posted by SE-user Hunter

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.