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

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.

• 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. – issoroloap May 18 '16 at 23:39
• @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 $T = \mathcal{P} \exp \int\limits_{-\infty}^{\infty} \mathrm{d} x \; U$? – Hunter May 18 '16 at 23:53
• Yes, you are right that the rapidly decreasing case and the periodic case deserve a somwhat separate treatment. I quote from "Integrable Systems. I" by B.A. Dubrovin, I.M. Krichever, S.P. Novikov, which you can find in "Dynamical Systems IV" and which I recommend as one of the best references: "In the periodic case, the spectral theory is completely different and bears no resemblance to the scattering theory". – issoroloap May 22 '16 at 10:10
• In a nutshell the monodromy matrix in the rapidly decreasing case (what DKN call the scattering theory) is the transition matrix between two basis of solutions to the spectral problem $L \psi = \lambda \psi$: the basis of solutions that are asymptotically periodic at $-\infty$ and $+\infty$, respectively. Such monodromy plays a similar role to what you defined for the periodic case, in particular its trace is a constant of motion. Anyway the rapidly decreasing and periodic cases are both treated in the reference I gave you. It's a really good read. – issoroloap May 22 '16 at 10:14
• @issoroloap thank you very much for the reference! I will start reading it immediately. – Hunter May 22 '16 at 11:58

One general idea here is that if we have the (generalized) Lax equations $$L(\psi)=0, \quad \psi_t=A(\psi),$$ where $L$ and $A$ are linear differential operators involving the spectral parameter $\lambda$ then the conservation laws, and hence the integrals of motion, for the associated integrable system can be obtained from the formal expansion of $\psi$ with respect to $\lambda$.