I am currently working on a project in acoustics and I am studying first and second-order approximations to the Navier-Stokes equation. I have been reading the book 'Theoretical Microfluidics' by Hendrick Bruus. You find the lecture notes corresponding to this book here. (Chapter 13: Acoustics in compressible liquids.)

The Navier-Stokes equation writes

$\rho \left[\frac{\partial \underline{u}}{\partial t} + (\underline{u}\cdot \nabla) \underline{u} \right]= -\nabla p + \eta \nabla^2 \underline{u} + \beta \eta \nabla (\nabla \cdot \underline{u}) \, .$

Assuming there is no "background flow" and using $p = p(\rho)$, the first-order fields are given as

$ \rho = \rho_0 + \rho_1 $

$ p = p_0 + c_a^2 \rho_1$

$ \underline{u} = 0 + \underline{u}_1$

where $\rho_0$ and $p_0$ are constants. When using first-order perturbation theory, all higher order terms, i.e. the product of two first-order terms, are neglected and the nonlinear term $(\underline{u}\cdot \nabla) \underline{u}$ is dropped. The linearized equation then is

$\rho_0\frac{\partial \underline{u}_1}{\partial t} = -c_a^2 \nabla \rho_1 + \eta \nabla^2 \underline{u}_1 + \beta \eta \nabla (\nabla \cdot \underline{u}_1) \, .$

My question is, how do we know that also the gradient of a first-order field, i.e. $\nabla \underline{u_1}$, is small and we therefore can drop the nonlinear term? Does this directly follow from the fact that $\underline{u}_1$ is small?

Thank you for your help.

  • $\begingroup$ The gradient is not necesarily small but the fact that it is multiplied by a small number makes it negligible compared to the other terms in the equation. So the assumption is that we can ignore it. $\endgroup$
    – nluigi
    Feb 7, 2019 at 15:12
  • $\begingroup$ Thanks for your response! There is one thing I still don't understand then. If we assume $\nabla \underline{u_1}$ is not small (i.e. order 0) then the term $\underline{u_1} \cdot (\nabla \underline{u_1})$ is of first-order and therefore not negligible. $\endgroup$ Feb 10, 2019 at 23:34
  • $\begingroup$ you misunderstand, $\nabla \underline{u_1}$ is first order just like all the terms in the final equation. $\underline{u_1} \cdot (\nabla \underline{u_1})$ is then second order and is negligible compared to the other terms. A decision if a term is negligible is always relative to the other terms. $\endgroup$
    – nluigi
    Feb 11, 2019 at 15:45

1 Answer 1


Sometimes these analyses are made easier if you use a parameter for tracking the orders. This is usually done in pertubation theory where a variable is expanded in terms of small parameter $\epsilon$, but the same idea can be applied here.

Take your equation: $$\rho \left[\frac{\partial \underline{u}}{\partial t} + (\underline{u}\cdot \nabla) \underline{u} \right]= -\nabla p + \eta \nabla^2 \underline{u} + \beta \eta \nabla (\nabla \cdot \underline{u})$$

and the variables expanded up to first-order terms: $$\rho = \rho_0 + \epsilon\rho_1 + O\left(\epsilon^2\right)$$ $$p = p_0 + \epsilon p_1 + O\left(\epsilon^2\right)$$ $$\underline{u} = 0 + \epsilon\underline{u}_1 + O\left(\epsilon^2\right)$$

Here the parameter $\epsilon$ is introduced just so we can track the orders as discussed in the intro; zeroth-order terms are $O(1)$, first-order terms are $O(\epsilon)$, second-order terms are $O(\epsilon^2)$, etc.

Substituting we get: $$\left(\rho_{0}+\epsilon\rho_{1}\right)\left[\epsilon\frac{\partial\underline{u_{1}}}{\partial t}+\epsilon^{2}(\underline{u_{1}}\cdot\nabla)\underline{u_{1}}\right]=-\nabla\left(p_{0}+\epsilon p_{1}\right)+\epsilon\eta\nabla^{2}\underline{u_{1}}+\epsilon\beta\eta\nabla(\nabla\cdot\underline{u_{1}})$$

Now let's group terms;

$O(1)$: The only term that remains is: $$\nabla p_{0}=0$$ This is clearly satisfied if $p_0$ is some constant background pressure.

$O(\epsilon)$: Expanding and multiplying out all the order parameters we get: $$\rho_{0}\frac{\partial\underline{u_{1}}}{\partial t}=-\nabla p_{1}+\eta\nabla^{2}\underline{u_{1}}+\beta\eta\nabla(\nabla\cdot\underline{u_{1}})$$

Clearly we see that the non-linear term is not taken into account because it is second-order and therefore negligible in the equation for first-order terms.

Second-order terms are not taken into account because they were not included in the expansion. These could be included but wouldn't change the conclusion.

  • $\begingroup$ Ok, I understand. So in short this means that we assume that the derivative of a first-order term (time-derivative, gradient, ...) is of first-order again. You expressed this mathematically by treating $\epsilon$ as a constant, i.e. $\nabla(\epsilon \underline{u}_1) = \epsilon \nabla \underline{u}_1$. $\endgroup$ Feb 13, 2019 at 0:24

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