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I'm trying to undestand the self-similarity as an invariance of a function under certain transformation. For example I think $$f(\lambda x)=\lambda^\epsilon f(x)$$ could be understood as a self-similarity like follows: a stretch of the $x$ variable is equivalent to a scale factor $\lambda^\epsilon$ over the whole profile $f(x)$. However it is not clear to me, when multiple variable are involved, if self-similarity could "couple" them together. For example $$u(x,t)=At^{-\alpha}f(\xi) \qquad \xi=xt^{-\beta}$$ is a self-similar function? In this case it is not true $$u(\lambda x,t)=\lambda^\gamma u(x,t)$$ nor $$u(x,\lambda t)=\lambda^\delta u(x,t)$$ but $$u(x,\lambda t)=\lambda^{-\beta}u(\frac{x}{\lambda^\beta},t)$$ still holds. So a stretch over the time is equivalent to a stretch of the profile AND of the $x$ variable.

Is there a general rule to check if a function is self similar? I looked up online but nothing sufficiently clear.

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  • $\begingroup$ How is this a physics question? $\endgroup$ – ACuriousMind Jan 14 '16 at 15:03
  • $\begingroup$ In fluid dynamics, such as heat propagation, or vortex destruction by viscosity, self-similarity is a useful tool to solve PDEs tranforming them into ODE through their invariant group. This is a different interpretation of the renormalization group. $\endgroup$ – Riccardo Buscicchio Jan 14 '16 at 15:17
  • $\begingroup$ That self-similarity may appear in physics doesn't make questions about self-similarity any more a physics question than the appearance of integrals makes generic questions about integrals physics questions. $\endgroup$ – ACuriousMind Jan 14 '16 at 15:18
  • $\begingroup$ As Dave showed below, understanding self-similarity could be easier if conveyed using its physical interpretation. Exactly the same way integrals, at a first stage, could be better understood using the notion of summing over small slices, just like you do in physics in many situations. In fact, this is the usual way to approach them at a basic level. $\endgroup$ – Riccardo Buscicchio Jan 14 '16 at 15:24
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What's going on is that the definition of the scaling variable $\xi=x^{-1/\beta} t$ (note: I prefer to parameterize it this way) defines how to resale the the two variables in a way that generates a scale transformation:

$$ u(\lambda^\beta x, \lambda t) = A \lambda^{-\alpha} t^{-\alpha} f( x^{-1/\beta} t ) = \lambda^{-\alpha} u(x,t) $$

by using different, but related, scale factors for $x,t$ they drop out of the $f(.)$ factor. One way to think of this is that you have a set of "families" of self similar systems, one for each $\xi_0$ value, and in terms of the $x,t$ domain, they lie along particular curves of the form $t = \xi_0 x^{1/\beta}$.

If you were just given a two parameter function function (rather than assuming that this form holds), the process for checking would be to first identify $\xi=x^{-1/\beta} t$, re-express the equation in terms of it, and see if everything can be expressed in terms of it, except for an overall $t^{-\alpha}$ factor.

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  • $\begingroup$ Thanks, clear explanation. Actually, the two parametrization are equivalent, by the mapping $\xi \rightarrow \xi^{-1/\beta}$. So, just to be clear to have understood it correctly, the self-similarity actually involves a simultaneous transformation of both variables. Right? $\endgroup$ – Riccardo Buscicchio Jan 14 '16 at 15:36
  • $\begingroup$ I know, I'm just used to seeing the exponent in the scaling variable "attached to" the opposite of the one out in front of scaling function. $\endgroup$ – Dave Jan 14 '16 at 15:37
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Continuing Dave's answer (and using his notation), a practical way to check for this kind of self-similarity is to draw a log-log plot of $u(x,t)$ as a function of $x$ for different values of $t$. If $u$ is self-similar, the different curves would all all have the same shape and would be related by translations.

To see this, define $X \equiv \log x$ and $T \equiv \log t$. With these variables, $$ u(x,t) = u(e^X, e^T) = e^{-\alpha T} f(e^{T-X/\beta}) \equiv e^{F(X-\beta T)-\alpha T}, $$ where the last equality defines the function $F$. Therefore, $$ U(X,T) \equiv \log u(e^X,e^T) = F(X-\beta T) - \alpha T. $$ This way you can in fact identify more general scaling functions of the form $u(x,t) = a(t) f\big(x b(t)\big)$ for functions $a(t)$ and $b(t)$ that are not necessarily power-laws.

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