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Consider the Bejamin Ono equation \begin{equation} \partial_t u + H\partial_{xx}u = u\,\partial_x u, \end{equation} where $u=u(x,t): \mathbb{R}\times\mathbb{R}\to\mathbb{R}$ is a real scalar field, and $H$ is a Hilbert transform defined by \begin{equation} \widehat{Hf}(\xi) = -i\:sgn(\xi)\hat{f}(\xi) \end{equation} Assume that $u$ has enough regularity and decays at infinity. Show that $\|u(t)\|_{L^2_x(\mathbb{R})} = \|u(0)\|_{L^2_x(\mathbb{R})}$.

Here is what I've done. \begin{equation} \partial_t \int_\mathbb{R} u^2 \,dx = 2\int_\mathbb{R} (-u H\partial_{xx}u + u^2\partial_x u) \,dx = 2\int_\mathbb{R} \partial_x u \, H\partial_x u \,dx \end{equation} because using integration by parts we get \begin{equation} \int_\mathbb{R} u^2 \partial_x u \,dx = -2\int_\mathbb{R} u^2 \partial_x u \,dx, \end{equation} which implies $\displaystyle \int_\mathbb{R} u^2 \partial_x u \,dx = 0$. However, I'm stuck at showing \begin{equation} \int_\mathbb{R} \partial_x u \, H\partial_x u \,dx = 0. \end{equation} I'm thinking to use some Fourier analysis like $\displaystyle \int_\mathbb{R} f\overline{g} \,dx = \int_\mathbb{R} \hat{f} \overline{\hat{g}} \,dx$, but I haven't got good results yet. Am I on the right track, and how can I do this problem?

Thank you.

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  • $\begingroup$ The Hilbert transform commutes with derivatives and a real function and its Hilbert transform are orthogonal. $\endgroup$ Mar 17, 2016 at 0:04
  • $\begingroup$ I understand that $H\partial_x u = \partial_x Hu$. For the second part, can you elaborate more? Thank you. $\endgroup$
    – dh16
    Mar 17, 2016 at 0:06
  • $\begingroup$ I edited my answer accordingly. $\endgroup$ Mar 17, 2016 at 0:18

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One can show that the Hilbert transform commutes with derivatives, i.e. $H\partial_x = \partial_x H$. Moreover, for a real valued function $v$, $\langle v,Hv\rangle = 0$, i.e. $\int_{\Bbb R}v(x)Hv(x)\,dx = 0$. This can be shown by employing Parseval's theorem for the Fourier transform. If $v$ is real, $Hv$ is real, so $Hv = \overline{Hv}$, giving

$$ \int_{\Bbb R} v(x)\overline{Hv(x)}\,dx = \int_{\Bbb R}\mathcal{F}v(\xi)\overline{\mathcal{F}Hv(\xi)}\,d\xi.$$

Note then that $\mathcal{F}H = -i\operatorname{sgn}(\cdot)\mathcal{F}$, so that

$$ \int_{\Bbb R} v(x)\overline{Hv(x)}\,dx = -i\int_{\Bbb R}\operatorname{sgn}(\xi) |\mathcal{F}v(\xi)|^2\,d\xi.$$

Since $v$ is real, the real part of $\mathcal{F}v$ is even and the odd part of $\mathcal{F}v$ is odd so that $|\mathcal{F}v|^2$ is the sum of two even functions. Coupled with the appearance of $\operatorname{sgn}(\xi)$, you have your result.

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  • $\begingroup$ Just a quick question, can we justify that $v$ is real implies $Hv$ is real. I remember that the Fourier transform of $v$ is real if $v$ is real and even, but we don't know the symmetry here. Thank you. $\endgroup$
    – dh16
    Mar 17, 2016 at 1:30
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    $\begingroup$ $Hv$ is just an integral (well principal value) of real-valued functions so it is real-valued. $\endgroup$ Mar 17, 2016 at 1:31
  • $\begingroup$ Sorry, I've just got it now; just writing our the integral as you said. Thank you. $\endgroup$
    – dh16
    Mar 17, 2016 at 1:35
  • $\begingroup$ You're very welcome :) $\endgroup$ Mar 17, 2016 at 1:40

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