Consider Burgers equation: \begin{equation} u_t+uu_x-\epsilon u_{xx}=0 \quad , \quad u(x,0)=g(x) \quad , \quad u = 0 \mbox{ when } |x| \mbox{ large}, \end{equation} where $u=u(x,t):\mathbb{R}\times\mathbb{R}_+ \to \mathbb{R}$, $\epsilon>0$, and $g\in L^2(\mathbb{R})$. Show that \begin{equation} \partial_t \left( \frac{1}{2}\int_\mathbb{R} u_x^2 \,dx \right) \leq \frac{||g||^2_{L^\infty}}{4\epsilon} \int_\mathbb{R} u_x^2 \,dx, \quad t\in[0,T] \end{equation} Assuming all terms are well-defined, and you can also assume more conditions if necessary.

Moreover, what is the relationship to energy dissipation? What does it mean if the inequality is shown?

Here is what I have done so far:

1) If $\displaystyle E(t) = \frac{1}{2} \int_\mathbb{R} u(x,t)^2 \,dx$, then $E'(t) \leq 0$.

2) The PDE for $u_x$ is $u_{tx} + uu_{xx} + u_x^2 - \epsilon u_{xxx} = 0.$

3) The equality satisfying $u_x$ is \begin{equation} \partial_t \left( \frac{1}{2}\int_\mathbb{R} u_x^2 \,dx \right) + \frac{1}{2}\int_\mathbb{R}u_x^3 \,dx + \epsilon\int_\mathbb{R} u_{xx}^2 \,dx = 0. \end{equation}

I don't know how to proceed next. I'm stuck when trying to let $g$ appear because all of my integrals and expressions involve $dx$. I need $u(x,0)$ to be somewhere. I am thinking to relate the equality with $E(t)$, but I don't know how.

Any help would be appreciated. Thank you.


1 Answer 1


Set $v=u_x$, then we have $$ -\int_\mathbb{R} v^3 =-\int_\mathbb{R} v^2 u_x = 2\int_\mathbb{R}v v_x u \leq 2\| vu \|_2 \| v_x\|_2, $$ where the last step is Cauchy-Schwarz. Now apply Young's inequality with $\varepsilon$ to obtain that the RHS is further dominated by $$ \varepsilon^2 \| vu\|_2^2 + \varepsilon^{-2}\|v_x\|_2^2\leq \varepsilon^2 \| u\|_\infty^2 \| v\|_2^2 + \varepsilon^{-2}\|v_x\|_2^2. $$ Setting $\varepsilon^{-2}=2\epsilon$ (with $\epsilon$ as in the question) and combining all these we get $$ -\frac{1}{2}\int_\mathbb{R} v^3 \leq \frac{1}{4\epsilon}\| u\|_\infty^2 \|v\|_2^2 + \epsilon \| v_x\|_2^2. $$ Now we use the maximum principle that says $\| u(t)\|_\infty\leq \| u(0)\|_\infty=\| g\|_\infty$. Substituting this in the equation you got in 3) gives then $$ \frac{d}{dt} \frac{1}{2} \| v\|_2^2 \leq \frac{\|g\|_\infty^2}{4\epsilon}\|v\|_2^2. $$


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