# how to solve this scaling equation

I am confronted with the following scaling equation (in Laplace domain if that matters):

$\tilde\psi(s) = a \tilde\psi(s/b)$

I know the answer is such that $\tilde\psi(s) \sim s^{\beta} K(s)$ where $\beta = \ln a/\ln b$ and I don't really care what K is. I'm not sure if the functions $\tilde\psi(s)$ are need to solve this but I do know them if they are necessary. How is this type of equation solved?

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This type of equation is not solved, it is piddled with. You have to understand the relations between the values it implies, together with constraints from continuity.

When you say about a function that \tilde\psi(s)= g(\tilde\psi(h(s))), you are saying that the value at x is related to the value at h(s). This means that it is related to the value at h(h(s)), and the study of the equation is related to the study of the iterations of the function h (both positive and negative iterations, so you need all the back images too), so that you know which quantities are independent. Then you have infinitely many equations on these quantities, and further constraints from differentiability/continuity if you wish to impose it.

In this case $h(s)=s/b$, $g(s) = as$, and the equation you give follows. The iteration of h(x) is very simple:

$$h(h(h(...\mathrm{n-nestings} ... h(x)...) = {x\over b^n}$$.

There is a nice trick for reducing the annoyance in the iterations, which is to switch from s to another logarithmic variable $x$ such that $s=e^x$. Then dividing $s$ by $b$ is shifting $x$ by $\log(b)$. The iterates of shifting is just adding an integer multiple of $\log(b)$, so you learn that

$$\psi(x) = {1\over a} \psi(x+\log b)$$

where $\psi$ is the same as $\tilde\psi$ except with an exponential relabling of the x-axis.

$$\psi(x) = \tilde\psi(e^x)$$

The point is that there is a periodicity condition, which states that the value at $x+n\log(b)$ for any integer n is uniquely determined to be ${1\over a^n}$ times the value at x. By rescaling x, you can make the period equal to 1.

The values of the function are therefore determined by the values between 0 and 1.

The values at some other x is determined by multiplying by ${1\over a^n}$ where n is the integer part of x. So that if you consider

$$\psi(x) = {1\over a^x} K(x)$$

Then $K$ is a function of x which must be periodic with period 1. In terms of the original variables, you find your ansatz, except you neglected to say the function $K$ is periodic as a function of x, with period $\log b$. Periodicity is a strange sort of log-periodicity in the space.

I should point out that this method only works for positive a,b and for positive s, so that the logarithm is the real line. I didn't bother with the other cases, because they are unlikely to be physics, but you might want to ask on math.

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thanks for the answer. The problem is tougher than I envisioned. I'm still working through what you wrote and I wasn't clear what your $f(x)$ is. Is that the same as $\psi$? Also it seems that s/b would shift x by -log b. – BeauGeste Apr 5 '12 at 18:34
@BeauGeste: Oops, sorry f is $\tilde\psi$, and it's really not as hard as all that--- I'm sorry for being obfuscatory--- it's just that you don't solve it by rote algebraic methods, but by thinking about the web of relations implied by the equations on all the values of f(x) considered as variables. If you are interested in such things, there are a few functional equations which show up in physics, most notably Feigenbaum's functional equation for universality in period doubling, and related intermittency one, plus a few others (Maxwell used one to find the Maxwell velocity distribution). – Ron Maimon Apr 5 '12 at 18:50
I fixed it to match your notation. – Ron Maimon Apr 5 '12 at 18:59
Sorry for being difficult but I am not able to get the ansatz from your final final equation. I am subbing in $x = log s - log b$. Is this type of scaling equation covered in any textbooks? – BeauGeste Apr 6 '12 at 23:22
@BeauGeste: Unfortunately no, but I can explain it, hopefully more clearly: the subbing in is not quite right--- x=log(s) and x-log(b) is log(s)-log(b). This means that if you use the x variable (which is just the same information as the s variable when s is restricted to be positive), dividing s by b corresponds to shifting x by log(b). If you want intuition, use log-paper and graph $y=s^b$ (it turns into a straight line). This might be covered on a how-to-read log-graphs webpage or a tutorial, it's not really a formal topic in a math book (you just pick it up by osmosis). – Ron Maimon Apr 7 '12 at 4:29