# What is the operational way of discovering scale invariance of differential equations?

Context The answer here by @Keenan Pepper gives an instance for what it means for an algebraic or trigonometric formula to be scale invariant. For quick reference, I quote his answer here but with a slightly changed notation. He essentially says that

The Euclidean distance formula $$a^2 + b^2 = c^2$$ is scale invariant since $$(\lambda a)^2 + (\lambda b)^2 = \lambda^2(a^2+b^2) = (\lambda c)^2.$$ In other words, the formula retains its form when all the variables are scaled by a constant multiplicative factor $$\lambda$$.

Question What would it mean for a differential equation to be (not to be) scale invariant?

When I say differential equations, I have very simple physics equations in mind such as $$\frac{d^2x}{dt^2}+\omega_0^2x=0,~ (\text{Undamped oscillation})\\ \frac{d^2x}{dt^2}+\gamma\frac{dx}{dt}+\omega_0^2x=0,~(\text{Damped oscillation})\\ \frac{d^2x}{dt^2}+\gamma\frac{dx}{dt}+\omega_0^2x=F_0\cos\omega t~ (\text{Damped, forced oscillation})$$ where $$\omega_0,\gamma,\omega,F_0$$ are all constants and the variables are $$x$$ and $$t$$. For instance, let us scale $$x\to \lambda x$$, and $$t\to \lambda t$$ so that the LHS becomes $$\frac{d^2(\lambda x)}{d(\lambda t)^2}+\omega_0^2(\lambda x)=\frac{1}{\lambda}\frac{d^2x}{dt^2}+\lambda\omega_0^2x\neq0$$ unless $$\lambda=\pm 1$$. So what's the conclusion? The first equation not scale invariant?

• You don't have to scale time necessarily. For example, if you take a solution $u(t)$ of equation for (un)damped oscillation, it's easy to see that $\lambda u(t)$ is also a solution of the same equation. This is a symmetry of a set of solutions: you pick one solution, apply some transformation and get a solution again. – Evgeny Apr 16 '19 at 11:43