Lie symmetry methods of differential equations.

Question

If we have a differential equation $$ y^{(n)} = H( x, y, y', ...y^{(n-1)}) . $$ And say we have its two infinitesimal generators admitted by this equation. What do I mean by the invariance of this equation with respect to these two infinitesimal generators and with respect two those two infinitesimal generators what happens to the equation ? For example if one of them is $ G_{1} = \frac {\partial}{\partial{y}}$ and the other is $ G_{2} = x\frac {\partial}{\partial{y}}$. What happens to the equation with respect to these infinitesimal generators?. Thanks for any sort of hints or help

Answer 1

Let us first consider a simplified setting. Assume that in the two-dimensional space $\mathbb{R}(x)\times\mathbb{R}(y)$ you have a manifold defined by an equation $$ \mathcal{M}(x,y) = 0\quad. $$ This manifold is said to be invariant under the action of the differential operator $G$ of the form $$ G = \xi(x,y)\dfrac{\partial}{\partial x} + \eta(x,y)\dfrac{\partial}{\partial y} \quad, $$ if the following is true: $$ \Bigl.G[\mathcal{M}(x,y)]\Bigr\rvert_{\mathcal{M}} = \left.\left(\xi(x,y)\dfrac{\partial \mathcal{M}(x,y)}{\partial x} + \eta(x,y)\dfrac{\partial \mathcal{M}(x,y)}{\partial y} \right)\right\rvert_{\mathcal{M}} = 0 \quad. $$ In the equation above we first 1) Act with the operator $G$ on the manifold $\mathcal{M}$, and then 2) Evaluate the resulting expression at the points of $\mathcal{M}$. Note that the expression $G[\mathcal{M}(x,y)]$ does not have to vanish trivially, it may happen that $G[\mathcal{M}(x,y)]\neq 0$.

Coming back now to the differential equation. Let us rewrite it in a more general form: $$ F(x,y,y',\ldots,y^{(n-1)},y^{(n)}) = 0\quad. $$ It can be understood as a manifold in the space $\mathbb{R}(x)\times\mathbb{R}(y)\times\mathbb{R}(y')\times\ldots\times\mathbb{R}(y^{(n)})$.

Since the line above contains not only $x$ and $y$, but also $y$'s derivatives up to the order $n$, it is not exactly the operator $G$ who should be applied to the line above. Instead, we need to introduce its $n$-th prolongation $\underset{n}{G}$. This will be operator of the form $$ \underset{n}{G} = G + \zeta_{(1)}\partial_{y'} + \ldots +\zeta_{(n)}\partial_{y^{(n)}} \quad. $$ See Peter Olver's book "Applications of Lie Groups to Differential Equations" for details on how to calculate the prolongations of operators.

In analogy with usual manifolds, the equation $F=0$ is said to be invariant under the action of $G$ (or, more precisely, its prolongation $\underset{n}{G})$ iff $$ \left.\underset{n}{G}[F(x,y,y',\ldots,y^{(n-1)},y^{(n)})]\right\rvert_F = 0 \quad. $$ Again, note that after applying $\underset{n}{G}$ to $F$ we evaluate the result at the points of $F$.

Sidenotes:

1. In certain (quite rare but still physical) cases the functions $\xi$ and $\eta$ depend not only on $x$ and $y$, but also on derivatives $y^{(k)}$. The most celebrated example of such a generalized symmetry is the Runge-Lenz vector.

   2. The knowledge of symmetries of the differential equation can be very helpful for integrating it. Good news is that finding them is a rather straightforward (but a bit technical and lengthy process). There's a very nice package for Mathematica which does this, it comes with Brian Cantwell's book "Introduction to Symmetry Analysis".

Answer 2

In general Lie symmetry methods for differential equations helps to construct complexe solution of DE from simple one , for example take the heat equation thanks to Lie theory we can construct the heat kernel from constant solutions