Dhrubaditya Mitra

Singularities of hyper-viscous Burgers equation

In simulations of Navier--Stokes equation with hyperviscosity it is observed that the intermediate range between dissipation range and the inertial range has a hump. This is called ``bottle-neck''. The explanation of this phenomenon might be along the following line. The inertial range has a power-law scaling. For pure viscosity the dissipation range is an exponential fall-off of the form $e^{-k \delta}$ where $\delta$ is the distance to the nearest singularity in the complex space. Here the intermediate range is a matching between the exponential and the decreasing power-law function and the matching happens smoothly. In presence of hyper-viscosity the nature of the complex space singularity will be changed. Hence the nature of the dissipation range will be changed and might have the form $k^{\beta}e^{-k \delta}$ where $\beta$ is a positive power. Then the matching problem becomes more difficult and we might expect a hump. This idea cannot be easily put to test in the Navier--Stokes case, but can be put to test in the case of Burgers equation in the following way.

• First we have to check whether the Burgers equation shows the ``bottle-neck'' effect or not. This can be easily confirmed by careful numerical spectral simulation of the Burgers equation with and without hyperviscosity.
• Next for non-hyperviscous Burgers equation we proceed in the following way. In the frame of a shock, throw-away the time-derivative. Then you have the equation \begin{equation} \partial_x(u^2) = \nu \partial_{xx} u \end{equation} which can be integrated once to obtain, \begin{equation} u^2 - a^2 = \nu \partial_x u \end{equation} which can again be integrated to yield \begin{equation} u(x) = a \tanh \left(\frac{x}{4\nu}\right) \end{equation} The Fourier transform of this function should give the correct form of the tail of the spectrum in Fourier space.
• To obtain the correct form of the spectrum in the presence of hyperviscosity is more complicated. We need to study the ordinary differential equation \begin{equation} \partial_x(u^2) = \nu \partial_{xx} u + \nu_h \partial_{xxxx} u \end{equation} on which one integration gives \begin{equation} u^2 - a^2 = \nu \partial_x u + \nu_h \partial_{xxx} u \end{equation} It is the third-derivative term which creates the problem. We might study the fixed points of this differential equation to obtain some insight into the problem. A way to tackle this equation in a perturbative way is the following. Rescale $x$ to transform $x \to x/\nu$, and $\epsilon = \nu_h/\nu^3$. Then we have \begin{equation} u^2 - a^2 = \partial_x u + \epsilon \partial^3_x u \label{u} \end{equation} Then start perturbation is $\epsilon$ by \begin{equation} u = u_0 + \epsilon u_1 + \epsilon^2 u_2 + \ldots \end{equation} Substituting this in Eq.~\ref{u} we get the following equation in different orders of $\epsilon$. \begin{eqnarray} u_0^2 - a^2 &=& \partial_x u_0 \hspace{1cm} {\rm (zeroth order)} \\ 2u_0u_1 &=& \partial_x u_1 + \partial_x^3 u_0 \hspace{1cm} {\rm (first order)} \\ 2u_0u_2 + u_1^2 &=& \partial_x u_2 + \partial_x^3 u_1 {\rm (second order) } \end{eqnarray} The zeroth order equation has the solution \begin{equation} u_0 = a \tanh\left(\frac{x}{4}\right) \end{equation} We next need to solve the first order equation.
• Otherwise we might use dominant ballance and Painleve analysis to study the nature of the complex space singularities of the hyperviscous equation. The cube-derivative would probably give a cubic pole in the complex plane which will contribute to $k^2exp(-\delta k)$ to the tail of the spectrum. This would illustrate the positive power of $k$ prefactor to the exponential and then we might try to match this with a power-law in the inertial range and gain insight into the problem.

Both the methods are somewhat hazy after the initial confidence but can perhaps be carried to conclusion.