Periodicity of KdV equation in relation to zero-curvature equation

Question

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

Comments

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

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@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