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I'm studying the viscous Burgers equation, $$ \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} = \nu \frac{\partial^2 u}{\partial x^2}, $$ and I came across this paper, that studies the equation with $\nu=1$ and infinite domain. In the paper, which defines the energy as $$ E(t) = \frac{1}{2} \int_\mathbb{R} |u(x,t)|^2 \exp \frac{x^2}{4} \, \mathrm{d} x, $$ I found the interesting energy bound (page 5) $$ E(t) \leq E(0) (t+1)^{-3/4}, $$ which is given without proof just after the definition of a weak solution for the equation.

The thing is, I am studying the equation with generic viscosity, and I tried comparing the energy decay in my (numerical) solution, using $$ E(t) \leq E(0) \left( \frac{t}{\nu} +1 \right)^{-3/4}, $$ and I verified that the energy and the bound are remarkably close for small viscosities and that the bound works for larger viscosities. My motivation for trying this was that changing $t$ for $\tau/\nu$ in the heat equation (a 'convection-less' form of the Burgers equation) would lead it from $$ \frac{\partial u}{\partial t} = \nu \frac{\partial^2 u}{\partial x^2} \ \text{ to } \ \frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial x^2}. $$

Now, I would like to understand how to obtain the energy bound, so I can try to work it with an arbitrary viscosity and justify the energy bound I 'found'. Any help would be appreciated.

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  • $\begingroup$ How is the energy defined? $\endgroup$ Commented Sep 12, 2020 at 22:17
  • $\begingroup$ @RobertLewis I added the definition of the energy. It is the 'classic' definition, except for the Gaussian weight function. $\endgroup$
    – rafa11111
    Commented Sep 12, 2020 at 22:44
  • $\begingroup$ Thanks. Are you sure you don't mean $e^{-x^2/4}$? It seems that $e^{x^2/4}$ might make things blow up too often too fast. $\endgroup$ Commented Sep 12, 2020 at 22:47
  • $\begingroup$ @RobertLewis It makes sense, but it's indeed $\mathrm{e}^{x^2/4}$ (equation 1.2 in the paper). The solution $u$ seems to satisfy $u\to 0$ as $|x| \to \infty$, but it needs then to decrease really fast, right? Anyway, I tried 'my' energy bound numerically without using any weight in my definition of energy, so I guess the bound does not depend on the norm one is using? $\endgroup$
    – rafa11111
    Commented Sep 12, 2020 at 22:55
  • $\begingroup$ I don't know answers to your questions, except that yes, $u \to 0$ really fast. $\endgroup$ Commented Sep 12, 2020 at 22:59

1 Answer 1

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You can scale $u$ too. Suppose $v(x,t)$ satisfies the Burgers equation with viscosity $\nu$, $v_t+vv_x-\nu v_{xx} = 0$. Define a new function $u(x,s)$ by $v(x,t) = au(x,s), s = bt$, with $a$ and $b$ to be determined. Substitute to see that $$0=v_t+vv_x-\nu v_{xx} = abu_s+a^2uu_x-\nu au_{xx}.$$ Now choose $a$ and $b$ so that $ab = a^2 = \nu a$, that is, $a = b = \nu$. Then $u$ satisfies the Burgers equation with $\nu = 1$. Now whatever has been proved for $\nu=1$ can be transferred to a statement about the general case by scaling $t$ and $u$ both.

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  • $\begingroup$ Thank you! That's what I ended up doing, but I also applied a scale factor to $x$. The 'remarkable' agreement I found was actually based on an error on my code so in a sense the premise of my question was flawed. The bound using the transformation you proposed worked quite well with the numerical solution, but it was more conservative than I would expect. $\endgroup$
    – rafa11111
    Commented Sep 30, 2020 at 13:33

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