9
$\begingroup$

The Burgers equation can be understood as a simplification of the Navier-Stokes equations when the pressure term is neglected:

$$ \frac{\partial u_i}{\partial t}+\ u_j\frac{\partial u_i}{\partial x_j}=-\frac{1}{\rho}\frac{\partial p}{\partial x_i}+\nu\frac{\partial^2u_i}{\partial x_j^2} \quad\Rightarrow \quad \frac{\partial u_i}{\partial t}+\ u_j\frac{\partial u_i}{\partial x_j}=\nu\frac{\partial^2u_i}{\partial x_j^2} $$

Using the Burgers equation instead of the Navier Stokes equation has the advantage that there is an analytical solution for the Burgers equation. This can be obtained using the Hopf-Cole transformation, which reduces the Burgers PDE, which is nonlinear, to the heat equation, which is linear.

However, when is it possible to neglect the pressure term $\boldsymbol{\nabla}p$ in the Navier-Stokes equations? What could be a physical example of such a situation?

$\endgroup$
0

3 Answers 3

2
$\begingroup$

Two straight-forward scenarios I can think of off the top of my head are:

  1. The flow is horizontal (probably also narrow?) such that the pressure gradient is sufficiently small as to be zero (i.e., $\nabla p\approx0$ as in Burgers' equation). A constant flow in a pipe would be one example here.
  2. The pressure gradient is identically equal to the external body forces (i.e., $\nabla p=\mathbf f$ such as in hydrostatics wherein $p=\rho gz,\mathbf f=\rho g\hat{z}$).

A third reason, of course, could be pedagogical in which a test might ask you to solve the Navier-Stokes equations without the pressure term to make the problem a little easier to solve given time constraints, but I think that's probably less a reason than you're wanting.

$\endgroup$
2
  • 1
    $\begingroup$ Not sure that constant flow in a pipe would be a good example. How would you get Hagen-Poisseuille? ;) $\endgroup$
    – kricheli
    Commented Jul 8, 2022 at 7:21
  • $\begingroup$ @kricheli To be fair, I claimed it exists as a scenario, not that it was interesting :D $\endgroup$
    – Kyle Kanos
    Commented Jul 8, 2022 at 12:52
2
+100
$\begingroup$

Two relevant examples, the decay of an initially linearly increasing wind and a source of water spreading out in a circular symmetric pattern, are illustrated here. As for neglecting the pressure gradient, one may think it as a toy model of the barotropic NS equations where pressure is a function of density alone , $p(\rho)=\rho^\gamma$ for $\gamma\rightarrow 0$. More interestingly, Burgers' equation also provides a simple model for Kolmogorov's turbulent flow having finite, nonvanishing dissipation as viscosity vanishes and the Reynolds number goes to infinity. In the limit the equation becomes hyperbolic and energy dissipation is concentrated in shock waves.

$\endgroup$
1
$\begingroup$

The pressure gradient is usually required in the direction of flow of a viscous fluid in order to overcome the effects of viscous dissipation. Moreover, the pressure gradient in the Navier-Stokes equations is associated with a constraint force whose sole role is to maintain the continuity constraint. In the Burgers’ equation, the absence of the pressure gradient implies the absence of the constraint force.

The Burgers' equation have a convective term, a diffusive term and a time-dependent term as follows: $$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=\nu\frac{\partial^2u}{\partial x^2}$$ The equation is parabolic when the viscous term is included and hyperbolic when the viscous term is neglected. An example of wave development for the viscous Burgers' equation is shown in the following figure:

1

The effect of the viscous term is twofold. First, it reduces the amplitude of the wave for increasing time. Second, it prevents multi-valued solutions from developing.

If the viscous term is dropped from the Burgers' equation the non-linearity allows discontinuous solutions to develop. The way this can occur is illustrated schematically in the following figure:

2

A wave is convecting from left to right and solutions for successive times are indicated. Points on the wave with larger values of $u$ convect faster and consequently overtake parts of the wave convecting with smaller values of $u$. It is necessary to postulate a shock across which $u$ change discontinuously to have a unique solution and so a physically result.

The above features make Burgers' equation a very suitable model for testing computational algorithms for flows where severe gradients or shocks are anticipated.

$\endgroup$
2
  • $\begingroup$ The question asked was, "When can the pressure term be dropped in NS?" not, "Why is Burger's equation interesting?" (or something of that sort). $\endgroup$
    – Kyle Kanos
    Commented Jul 8, 2022 at 12:58
  • $\begingroup$ The first paragraph is a direct attempt at answering the question. I think it's perfectly valid. $\endgroup$
    – Donn
    Commented Jul 13, 2022 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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