# Non-dimensionalization and perturbative expansion

I need to expand an equation, of the form $$\dot{r} = \gamma(a,\mu) F_1 + g(\mu,\ell,h,R) F_2$$

in powers of $$\epsilon = a/\ell$$.

So I thoughts I would non-dimensionalize it first.

I know that $$\gamma(a,\mu) = 1/6\pi a \mu,$$ with $$\mu$$ a dynamic viscosity, $$a$$ a length scale. $$g$$ is a complicated function of $$\mu$$ and $$\ell,h,R$$ which are also length scales. $$F_{1,2}$$ are forces, $$\dot{r}$$ is a velocity.

I would like to measure lengths in units of $$\ell$$, and times in units of $$\Omega^{-1}$$, where $$\Omega$$ is a frequency.

It is not clear to me how can I get to a position where I can perform an expansion in $$\epsilon$$.

• It seems somewhat straight forward to write the equation in dimensionless form. What have you tried ? – Adam Jan 24 at 16:33
• The problem arises when I have terms only in $a$, such as in $\gamma$. First of all, that does not give me $a/l$. Should I substitute $a=\epsilon \ell$? Also, I would get $1/\epsilon$, which is not ideal if I am to expand in small $\epsilon$. – usumdelphini Jan 24 at 19:49
• If you write explicitly the dimensions of every term (in length and time and ...), it will be easy to sort this out. For example, by dimensional analysis, you can write $g(\mu,l,h,R)=l^{\alpha}\Omega^{\beta} g(\mu l^{\gamma} \Omega^{\delta},1,h/l,R/l)$, etc. This should then make the rest easy. – Adam Jan 24 at 20:02

You need to measure quantities in units of $$\ell,\Omega$$, and μ, so, then, viscosity units instead of mass--which has units of $$\mu\ell/\Omega$$.

The left-hand-side has units of $$\ell \Omega$$, which must also agree with the units of the right-hand side.

The forces F have units of $$\mu \Omega \ell^2$$; and $$g$$ has the same units as $$\gamma$$, namely $$1/\ell \mu$$.

As a consequence, in these units, the right-hand side reads, for dimensionless $$f\equiv F/ \mu \Omega \ell^2$$, $$\ell \Omega\left (\frac{f_1}{6\pi \epsilon } + g\mu \ell ~f_2 \right ).$$

The parenthesis now consists of dimensionless quantities. As discussed, the dependence of the otherwise complicated function g on $$\mu$$ is simple, given its over-all dimension, $$g(\mu,\ell,h,R)=\tilde g (\ell,h,R) /\mu ,$$ where $$\tilde g$$ still has units of $$1/\ell$$, which can be trivially scaled out of all of its length variables, as indicated below.

Defining $$\tilde h\equiv h/\ell$$ and $$\tilde R\equiv R/\ell$$, you end up with a dimensionless r.h.-side parenthesis, $$\left (\frac{f_1}{6\pi \epsilon } + \tilde g(1,\tilde h, \tilde R) ~f_2 \right ).$$

You might experiment with simple notional example functions to appreciate the logic of the scaling.

If, indeed, you need to expand in terms of ε and not 1/ε, the form of the equation is unforgiving, and multiplying the equation by ε merely dictates excessive dominance of the of the first term to zeroth order, leading to its vanishing. But this non-dimensionalization is unique, given your desiderata. Perhaps you wish to modify them by, e.g., measuring lengths in units of a, instead.

Edit on alternate length units as per comment : If you wanted, instead, to use a instead of $$\ell$$ as your unit length, $$\epsilon$$ would disappear from the first term, and would only live on the second term. The second term might be proportional to $$\epsilon$$, depending on the explicit form of g. Recall g must have dimensions of 1/μ×length; so, e.g., for a purely hypothetical toy original
$$g= \frac{ 1}{\mu \ell} ( 1+ (h/R)^2 + h/\ell +...)$$ the corresponding nondimensionalized function multiplying the nondimensionalized second force $$F_2$$ would be $$a\mu g =\epsilon (1+(h/R)^2 + \epsilon (h/a)+...) .$$ and then expansion in $$\epsilon$$ would be regular... But without knowledge of specifics, one really cannot say much...

• Could you please show how measuring lenghts in terms of $a$ would change the leading term in $\epsilon$? Would this be constant instead? – usumdelphini Feb 6 at 9:13
• If you used a instead of $\ell$ in your units, $\epsilon$ would disappear from the first term, and the second term would have a dimensionless $g'(1/\epsilon, h/a,R/a)$ instead of the $\tilde g$ above. If the leading behavior of g were 1/$\ell$, the second term would go like $\epsilon$. You might have to give me a toy exemplary g to illustrate.... – Cosmas Zachos Feb 6 at 14:45