Symmetries and invariants.

Question

I would like someone to give me an example of application of the following paragraph from an introductory book on Lie groups theory:

The paragraph states the following:

Symmetry transformations, if they exist, play an important role in the construction of solutions of differential equations. Indeed, they make it possible to find differential invariants directly under the action of a group of symmetry ( Lie group ). These differential invariants decrease the order or degree of the differential equation in question. Thus, the discovery of more symmetries leads to more differential invariants, resulting in a decrease in the order or degree of the equation, which finally allows to find its solutions.

Thanks in advance for your help.

Answer 1

This is used a lot in physics; see Noether's theorem and constant of motion, in particular integral of motion.

An example is afforded by the conserved energy of a physical system. For instance, the equation of motion for a system consisting of a mass and spring is

$$ m\ddot x+kx=0\;, $$

a second-order differential equation. Multiplying both sides by $\dot x$ and integrating over time yields

$$ \frac12m\dot x^2+\frac12kx^2=E\;, $$

where the free additive constant $E$, the energy, consisting of the kinetic energy (the first term) and the potential energy (the second term), is a first integral of the motion, and we now have a first-order differential equation. We've introduced an additional constant and reduced the order of the differential equation by one.

The symmetry related to this constant of the motion is the invariance of the system (and thus of the equation of motion) under time translations. Similarly, invariance under spatial translations leads to conservation of linear momentum, and invariance under rotations leads to conservation of angular momentum.

Answer 2

Suppose you have the ODE $$y^{\prime\prime}(x) = f(x,y(x),y^\prime(x))$$ and suppose that someone told you that the Lie group $\bar y = y+\epsilon$ is a symmetry of the ODE. That means that $$ \bar y^{\prime\prime}(x) = f(x,\bar y(x)-\epsilon,\bar y^\prime(x)) = f(x,\bar y(x),\bar y^\prime(x)).$$ But this means that $f(x,y,z) = f(x,z)$, that is the ODE has the form $$y^{\prime\prime}(x) = f(x,y^\prime(x)).$$ Now by saying that $y^\prime(x)=p(x)$ we get the reduced ODE $$p^\prime(x) = f(x,p(x)).$$