The viscous Burgers' equation: $$ q_{t}+q\:q_{x}~=~\nu\:q_{xx}, \mbox{ where } \:\:\nu >0, $$ combines the nonlinear propagation of $q(x,t)$ and the diffusion.

  • What is this equation for? (in modelling context)

  • What is so special about it? (other than if $\nu=0$ then the solution can form a shockwave)


2 Answers 2


1D Burger's equation is not meant to model a physical phenomenon. Rather, it is a simplification of homogeneous incompressible Navier-Stokes equations that preserves (some of) its mathematical structure: the non-linear convection term and the second order derivative of viscous forces.

It was initially intended as a useful simplification to try to understand the mathematical problems present in N-S equations, e.g. the limit of small viscosities (high Reynolds number). It is not anymore a tool for researchers studying existence and uniqueness of solutions of N-S equations, but is still useful to test numerical schemes designed for N-S.

Its multidimensional version also arises in some other physical contexts, see http://arxiv.org/abs/nlin/0012033

  • 2
    $\begingroup$ but it's still useful to test numerical schemes pretty much how Randy LeVeque treats it. Also useful for instructing someone on discretion methods of hydrodynamic, which is related to your comment. $\endgroup$
    – Kyle Kanos
    Jul 23, 2014 at 11:04
  • $\begingroup$ I think it also provides one of the ways the Reynolds number is derived, or at least physically demonstrated. For the form shown in the OP, R = $\frac{ q \ q_{x} }{ \nu \ q_{xx} }$, which shows that the Reynolds number can represent the ratio of steepening to viscosity (or in KdV equations, viscosity is replaced by dispersion). $\endgroup$ Oct 31, 2014 at 18:17

The convection-advection equation resembles the mathematical structure of Burgers equation. Despite differences (convection-advection has an "extra" variable $c(x,t)$) you can think that Burgers describes compressible material fluid transportation by means of advection (the $ \vec{v} \cdot \nabla $ term) and diffusion (the $D \nabla^2$ term) without mechanical pressure gradients applied.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.