Invariant of KdV equations

Question

This is something I saw in a paper. Let me describe it below:

Consider the KdV equation on $(0,1)$ subject to the periodic boundary conditions for all derivatives: \begin{equation} \begin{aligned} \frac{d}{dt}u + D^3 u + u D u & = 0 \\ u(0) & = u_0 \end{aligned} \end{equation} where $D$ denotes the spacial derivative. We consider following invariant of the equation: \begin{equation} \begin{aligned} L(v) := \int_0^1 (D^2 v)^2 - \frac{5}{3} v(Dv)^2 + \frac{5}{36}v^4. \end{aligned} \end{equation} We could compute the Frechet derivative for $L$ \begin{equation} \begin{aligned} \langle L'(v), h \rangle := \int_0^1 2(D^2v)(D^2h) - \frac{5}{3}(Dv)^2 h - \frac{5}{3} D(v^2)Dh + \frac{5}{9}v^3h \end{aligned} \end{equation} Then it is claimed that $\langle L'(u), D^3 u + u D u \rangle = 0$, where $u$ solves the KdV equation.

I am not able to prove this fact. It should follow from integration by parts, but I failed to play with those many terms. I think one property for $L$ is that $\langle L(u), D^3 u + u D u \rangle = 0$, could we derive $\langle L'(u), D^3 u + u D u \rangle = 0$ from here? Could anyone do a bit of computation for me?

Edit: We could assume everything is smooth at this moment.

Answer 1

Assume that $u$ is a smooth solution of the KdV equation. We denote $h=D^3u+uDu$ , $$I_1=\int_0^1D^2uD^2h,\ \ I_2=\int_0^1(Du)^2h,\ \ I_3=\int_0^1u(Du)(Dh),\ \ I_4=\int_0^1u^3h, $$ and $$A=\int_0^1Du(D^2u)^2,\qquad B=\int_0^1u(Du)^3.$$ Then $$\langle L'(u), h\rangle=2I_1-\frac53I_2-\frac{10}3I_3+\frac59I_4.$$

In the following we prove that $\langle L'(u), h\rangle=0$.

Notation: $\int=\int_0^1$.

We have $I_1=\int D^2u(D^5u+D^2(uDu))$. Since $$\int D^2uD^5u=-\int D^3uD^4u=-\int D^3uD(D^3u)=-\frac12\int D((D^3u)^2)=0,$$ we have \begin{align*} I_1&=\int D^2u D^2(uDu)=\int D^2u(3Du D^2u+u D^3u)=3\int Du(D^2u)^2+\int uD^2uD^3u\\ &=3\int Du(D^2u)^2+\frac12\int u D((D^2u)^2)=\frac52\int Du(D^2u)^2=\frac52A. \end{align*} For $I_2$, we have \begin{align*} I_2&=\int (Du)^2D^3u+\int u(Du)^3=\int (Du)^2D(D^2u)+B\\ &=-2\int Du(D^2u)^2+B=-2A+B. \end{align*} For $I_3$, we have \begin{align*} I_3&=\int u Du D^4u+\int u Du D(u Du)=\int D^2(u Du)D^2u+\frac12\int D((u Du)^2)\\ &=I_1+0=\frac52A. \end{align*} For $I_4$, we have \begin{align*} I_4&=\int u^3D^3u+\int u^4 Du=\int D^2(u^3)Du+\frac15\int D(u^5)\\ &=\int 6u(Du)^3+3u^2D^2uDu=6B+\frac32\int u^2D((Du)^2)\\ &=6B-3\int u(Du)^3=3B. \end{align*} Therefore, $\langle L'(u), h\rangle=0$.