3

Bourgain proved that the periodic KdV equation $$\begin{align} \partial_t u+\partial_x^3 u+u\partial_x u&=0\\u(0,x)&=u_0(x)\end{align}$$ is locally well-posed in $H^s(\mathbb T)$ in [1]. Here is the sketch: define the $X^{s,b}$ space as the closure of the set of Schwartz functions $\mathcal S_{t,x}(\mathbb R\times\mathbb T)$ under the norm $$\|u\|_{X^{s,b}}:=\|\langle k\rangle^s\langle\tau-k^3\rangle^b\hat u(\tau,\xi)\|_{L_\tau^2l_k^2}$$ for $\{s,b\}\subset\mathbb R$ with $\langle\cdot\rangle:=1+|\cdot|$. Consider a transformation $T$: $$Tu=\eta(t)e^{Lt}u_0-\eta(t)\int_0^te^{L(t-t')}(u\partial_xu)dt'$$ where $L:=-\partial_x^3$ and $\eta$ is a suitable smooth cutoff. Then one can show the transformation $T$ is a contraction map on a bounded subset of $X^{s,1/2}$ and thus yields a unique fixed point $Tu=u$ in $X^{s,1/2}$. Therefore by Duhamel's principle, the periodic KdV equation has a unique local solution $u\in X_I^{s,1/2}$ ($X^{s,1/2}$ but restricted to a short time interval $I$).

Here is my question: how does this even prove the well-posedness in $H^s$? As far as I know there is no inclusion relation between $X_I^{s,1/2}$ and $L_{t,\text{loc}}^\infty H_x^s$. In fact one can show the $X_I^{s,1/2}$ solution is in $L_{t,\text{loc}}^\infty H_x^s$, but that is not obvious from his argument, and most of all I don't see why there can't be another solution that is not in $X_I^{s,1/2}$. What am I missing? Below is what I tried to prove the uniqueness of the solution with $s=0$.

Assume $u,v\in C(I,L^2(\mathbb T))$ are both solutions for the same initial data $u_0$ and let $w:=u-v$. Then $w$ satisfies $\partial_tw+\partial_x^3w+w\partial_xu+v\partial_xw=0$. Multiplying both sides by $w$ and integrating in space yields $$ \frac12\int_\mathbb T\partial_t(w^2)dx+\int_\mathbb Tw\partial_x^3wdx+\int_\mathbb Tw^2\partial_xudx+\int_\mathbb Twv\partial_xwdx=0. $$ Using $2wv\partial_xw=\partial_x(vw^2)-w^2\partial_xv$ and integrating by parts, we then have $$\frac12\int_\mathbb T\partial_t(w^2)dx-\frac12\int_\mathbb T\partial_x((\partial_xw)^2)dx+\int_\mathbb Tw^2\partial_xudx+\frac12\int_\mathbb T\partial_x(vw^2)dx-\frac12\int_\mathbb Tw^2\partial_xvdx=0.$$ It follows then that $$\frac12{\frac d{dt}}\|w(t)\|^2_{L_x^2}=-\int_\mathbb Tw^2\partial_xudx+\frac12\int_\mathbb Tw^2\partial_xvdx.$$ By Hölder, this implies $$\frac d{dt}\|w(t)\|^2_{L^2_x}\le2\left(\|u_x(t)\|_{L_x^\infty}+\|v_x(t)\|_{L_x^{\infty}}\right)\|w(t)\|^2_{L_x^2}$$ I want to use Grönwall's inequality, but there is no guarantee that $\|u_x(t)\|_{L_x^\infty}<\infty$.

[1] J. Bourgain, Fourier transform restriction phenomena for certain lattice subsets and applications to nonlinear evolution equations, Part II: The KDV-Equation, Geom. Funct. Anal. 3 (1993), 209-262.

Jaeseop Ahn
  • 679

1 Answers1

1

Bourgain's trick was to consider the following modified space $Y^{s}$ which embeds in $C_{t}H^{s}_{x}$: \begin{equation*} ||u||_{Y^{s}}=||u||_{X^{s,\frac{1}{2}}}+||〈k〉^{s}\hat{u}(\tau,k)||_{{\ell}^{2}_{k}L^{1}_{\tau}} \end{equation*} Then you can run the usual contraction mapping argument in this space.

  • I am aware, which is exactly why I mentioned that it is possible to show the $X_I^{s,1/2}$ solution is in $L_{t,\text{loc}}^\infty H_x^s$. And I am saying this does not eliminate the possibility there is a solution outside $Y^s$. – Jaeseop Ahn May 07 '22 at 18:28
  • That is called the 'unconditional well-posedness'. Bourgain's method only gives the conditional well-posedness, which means that the uniqueness depends on the auxiliary Banach space $Y^{s} \hookrightarrow C_{t}H^{s}$. To prove the unconditional well-posedness, you will need some different methods, for example differentiation by parts. –  May 09 '22 at 15:35
  • 1
    That was the answer I was looking for! Can you add that to your answer so that I can accept it? – Jaeseop Ahn May 21 '22 at 21:23