How does Lie group help to solve ode's?

Stephen Tashi

While we wait to see if anybody cares, could you explain whether your mathematical background includes directional derivatives and differential operators?

Currently, I can't answer the question ("In geometric terms, how does Lie theory helps solve differential equations?") and I only get a hazy idea from the post that said that it amounts to changing coordinates so every transformation amounts to a translation. I'll flatter myself by thinking that I am capable of getting geometric picture if I wade through some of the books that I have. (I like the Gilmore text.) However, I probably don't have the self discipline to do this without some motivation ( being a very busy person - retired, you know. It takes up all your time.) We might figure this out if you want to have a long discussion about it, which will keep me interested.

If you were just asking a casual question and wanted a quick answer without getting into a lot of posts, I can't help you.

jojo12345

Here is a geometric picture of how symmetries can aid in the solution of certain ODEs. I'll begin with some technical discussion and conclude with some qualitative remarks.

Let [itex] M [/itex] denote the phase space. This is the manifold where solution curves to the ODE are contained. Let [itex]X:M\rightarrow TM[/itex] denote the vector field that defines the ODE. A geometric way of saying that this ODE possesses symmetry is [itex] M [/itex] admits a (left) Lie group action [itex]\Phi:G\times M\rightarrow M[/itex] that commutes with the flow map of [itex]X[/itex]. Recall that the flow map [itex]F_\lambda[/itex] of [itex]X[/itex] takes a point [itex]m\in M[/itex] and returns the point [itex]F_\lambda(m)[/itex] equal to where the initial condition [itex]m[/itex] ends up after moving [itex]\lambda[/itex] seconds according to the dynamics defined by the ODE. So, restated in symbols, the condition that the ODE possess a symmetry is
[tex]\forall g\in G,\lambda\in\mathbb{R},~~\Phi_g\circ F_\lambda=F_\lambda\circ\Phi_g.[/tex]
Provided the action [itex]\Phi[/itex] is sufficiently nice (e.g., free and proper), then it is possible to form the quotient manifold [itex]M/G[/itex], which is just the collection of group orbits (see http://en.wikipedia.org/wiki/Group_action for more discussion on group actions, particularly the section on continuous group actions). On this smaller manifold [itex]M/G[/itex], the original ODE induces a new ODE [itex]x:M/G\rightarrow T(M/G)[/itex] via the following construction.

Let [itex]\pi:M\rightarrow M/G[/itex] denote the map that assigns to every point [itex]m\in M[/itex] the group orbit [itex]m[/itex] lives on. For each [itex]\lambda\in\mathbb{R}[/itex], define the mapping [itex]f_\lambda:M/G\rightarrow M/G[/itex] by [tex]f_\lambda(\pi(m))=\pi(F_\lambda(m)).[/tex]
This is a well-defined mapping precisely because the symmetry condition holds. That is, if [itex]m^\prime=\Phi_g(m)[/itex] lives on the same group orbit as [itex]m[/itex], then
[tex]
f_\lambda(\pi(m^\prime))=\pi(F_\lambda(\Phi_g(m)))=\pi(F_\lambda(m))=f_ \lambda(\pi(m)).
[/tex]
The key property this map possesses is that it defines a one-parameter subgroup of [itex]M/G[/itex], which in symbols means [itex]f_{a+b}=f_a\circ f_b[/itex] for any real numbers [itex]a,b[/itex]. Thus, [itex]f_\lambda[/itex] is the flow map of some vector field [itex]x[/itex] on [itex]M/G[/itex].

The vector field [itex]x:M/G\rightarrow T(M/G)[/itex] contains all of the "essential" information contained in the original vector field [itex]X:M\rightarrow TM[/itex]. In fact, if it turns out to be possible to solve the ODE defined by [itex]x[/itex], then the solution to the original ODE can be explicitly reconstructed in terms of integrals of the solution to [itex]x[/itex] (I won't explain how this works here). Note that because [itex]x[/itex] is defined on a space with smaller dimension than [itex]M[/itex] (in fact [itex]\text{dim}G[/itex] dimensions smaller), it is reasonable to expect that [itex]x[/itex] is easier to solve than [itex]X[/itex].

Finally, I'd like to point out that symmetry is even more useful when your ODE is Hamiltonian. In the Hamiltonian case, it is often possible to reduce the dimensionality of the system by TWICE the dimension of [itex]G[/itex]! This is because Hamiltonian symmetries also give rise to conserved momentum maps. An authoritative source for learning about Hamiltonian symmetries is Abraham and Marsden's Foundations of Mechanics.

algebrat

I believe symmetry can be use to explain many of the tricks from the chapter in Boyce on differential equations.

[itex](x,y)\mapsto(x,y/x)[/itex], homogeneous change of variables.

So we're to notice the lie group of symmetries along any y=mx. If we travel from [itex](x,y)[/itex] to [itex](x',y')[/itex], but stay on the line, ie y'=mx', then if y'=F(y/x), a certain symmetry arises.

Since [itex]dy/dx(x,y)=f(x,y=F(y/x)=F(m)=F(y'/x')=f(x',y')=dy/dx(x',y')[/itex]

So the theory of lie groups says the differential equation involving two variables splits into two of fewer variables if we use the new coordinates, instead of [itex](x,y)[/itex], we sort of use [itex](x,\tan(\theta))[/itex], so we move only radially and in [itex]x[/itex], and our shape simplifies.

Other changes in path due to some type of symmetry,

[itex](x,y)\mapsto(x,\mu(x)y)[/itex], the integrating factor. Splits the differential equation for a certain [itex]\mu(x)[/itex].

[itex](x,y)\mapsto (x,x+y)[/itex], if [itex]y'=f(x,y)[/itex] has symmetry [itex]y'=F(x+y)[/itex]

It would be an exercise to see what corresponds with what from an advanced text. Singer has a couple good books on symmetry and mechanics that are hiding some intuitions too. I haven't visited this topic in a while, so I just remember the parts that were relevant to me more often. Picking symmetries in calculus of variations is another related topic, giving nice "first integrals", which I think relates to energy conservation. Singer's correspondence between symmetries and conserved quantities in her book on Hydrogen makes it seem really simple, but it takes some mathematical buildup. The main idea was something like if an operator commutes with a transformation.

I think these are all good things to try to take into account to create a nice picture of the topic.