# Conservation law for Benjamin Ono equation

Consider the Bejamin Ono equation $$\partial_t u + H\partial_{xx}u = u\,\partial_x u,$$ 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 $$\widehat{Hf}(\xi) = -i\:sgn(\xi)\hat{f}(\xi)$$ 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. $$\partial_t \int_\mathbb{R} u^2 \,dx = 2\int_\mathbb{R} (-u H\partial_{xx}u + u^2\partial_x u) \,dx = 2\int_\mathbb{R} \partial_x u \, H\partial_x u \,dx$$ because using integration by parts we get $$\int_\mathbb{R} u^2 \partial_x u \,dx = -2\int_\mathbb{R} u^2 \partial_x u \,dx,$$ which implies $\displaystyle \int_\mathbb{R} u^2 \partial_x u \,dx = 0$. However, I'm stuck at showing $$\int_\mathbb{R} \partial_x u \, H\partial_x u \,dx = 0.$$ 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.

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

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.

• 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.
– dh16
Mar 17, 2016 at 1:30
• $Hv$ is just an integral (well principal value) of real-valued functions so it is real-valued. Mar 17, 2016 at 1:31
• Sorry, I've just got it now; just writing our the integral as you said. Thank you.
– dh16
Mar 17, 2016 at 1:35
• You're very welcome :) Mar 17, 2016 at 1:40