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 '19 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. $\endgroup$ – FrankHauser Feb 10 '19 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 '19 at 15:45

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. $\nabla(\epsilon \underline{u}_1) = \epsilon \nabla \underline{u}_1$. $\endgroup$ – FrankHauser Feb 13 '19 at 0:24

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.