1
$\begingroup$

The Buckingham-pi theorem says that a dimensional quantity of the form

$p = f(p_1, \cdots ,p_k,q_1 \cdots, q_n )$

(where the $p_i$'s dimensions form the fundamental set of units) can be rescaled as

$\tilde{p} = f(1,\cdots ,1,\tilde{q}_1, \cdots \tilde{q}_n)$

where $\tilde{p} , \tilde{q_i}$ are dimensionless.

The application of this to PDE's is throwing me off

For example in Navier-Stokes we have that a solution $u = u(x,t, \nu, p_0, L, u_0(x))$, the solution depends on position, time, viscosity, density, a length scale, and your initial velocity.

Clearly $L, u_0(x), p_0$ form a fundamental set of units for mass, length, and time.

This suggests that $u$ should be a function of 6-3 =3 variables , but upon nondimensionalizing the PDE we find $\tilde{u} = \tilde{u}(\tilde{x},\tilde{t}, \tilde{u_0}(x), Re)$, a function of four variables.

Somehow I am applying the Buckingham-pi theorem wrong, but I can't see how.

The only thing I can think of is that most things I've read factor in some sort of "typical velocity scale", but this should be dependent on $u_0(x)$. How can we identify any other velocity scale before we have solved a problem?

Is using an "infinite dimensional" variable like $u_0(x)$ throwing this off? If so why does $u_o(x)$ being infinite dimensional cause a problem but not $x$ or $t$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

I think the count is off, because the unit of mass does not appear as an independent degree of freedom in the Navier-Stokes equation (unless you include gravitational effects). You have the independent units of length $L$ and time $T$, and you use these two units to eliminate two of the seven variables (counting pressure $p$ and mass density $\rho$ separately). Specifically, following your notation, you start with a velocity field $u=u(x,t,\nu,\rho,p_0,L,u_0)$ and use $L$ and $T$ to arrive at a dimensionless solution $\tilde{u}=\tilde{u}(\tilde{x},\tilde{t},\tilde{p}_0,\tilde{u}_0(\tilde{x}),R)$, with Reynolds number $R=L^2/T\nu$. Having an independent unit of mass does not help you to reduce this further.

see these notes, page 99, for a more elaborate discussion how this works.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ I think I figured it out, the issue is that typically when people nondimensionalize NS they compare two similar scenarios i.e. a certain kind of flow in a certain geometry. Changing the initial condition $u_0(x)$ is too drastic of a change for us to expect it to be modeled by a single parameter ($Re$) family of solutions. How the problem should really be looked at is $u(\rho,\nu, U, L) \rightarrow \tilde{u}(Re)$ in which case 4-3 = 1 and Buckinham pi makes sense. $\endgroup$
    – Ryan
    Commented May 7, 2014 at 9:07
  • 1
    $\begingroup$ I guess the point is that when people nondimensionalize Navier Stokes, they neglect to mention they are nondimensionalizing it in a specific geometry with a specific flow (which allows you to consistently define $L$ and $U$). Varying an infinite dimensional parameter like $u_0(x)$ leads to trouble when using the Buckinham-pi theorem, although it's still not clear to me why. $\endgroup$
    – Ryan
    Commented May 7, 2014 at 9:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .