1.1 Terminology used and high level introduction

1. \(x,y\) are the natural coordinates used in the input ode \(\frac {dy}{dx}=\omega \left ( x,y\right ) \).
2. \(\bar {x},\bar {y}\) are called the Lie group (local) transformation coordinates. The ode remains invariant (same shape) when written in \(\bar {x},\bar {y}\). The coordinates \(R,S\) (some books use lower case \(r,s)\) are called the canonical coordinates in which the input ode becomes a quadrature and therefore easily solved by just integration.
3. \(\xi ,\eta \) are called the Lie infinitesimals. \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) can be calculated knowing \(\bar {x},\bar {y}\). Also \(\bar {x},\bar {y}\) can be calculated given \(\xi ,\eta \). It is \(\xi ,\eta \) which are the most important quantities that need to be determined in order to find the canonical coordinates \(R,S\). These quantities are called the tangent vectors. These specify how the orbit moves. The orbit is the path the point \(\left ( x,y\right ) \) point travels on as it move toward \(\bar {x},\bar {y}\). The tangent vectors \(\xi ,\eta \) are calculated at \(\epsilon =0\). The point \(\bar {x}=x+\xi \epsilon \) and the point \(\bar {y}=y+\eta \epsilon \).
4. The ultimate goal is write \(\frac {dy}{dx}=\omega \left ( x,y\right ) \) in \(R,S\) coordinates where it is solved by integration only as it will have the form \(\frac {dS}{dR}=F\left ( R\right ) \). The right hand side should always be a function of \(R\) only in canonical coordinates.
5. \(\bar {x},\bar {y}\) can be calculated knowing the canonical coordinates \(R,S\).
6. The ideal transformation has the form \(\left ( \bar {x},\bar {y}\right ) \rightarrow \left ( x,y+\epsilon \right ) \) because with this transformation the ode becomes quadrature in the transformed coordinates. But because not all ode’s have this transformation available, the ode is transformed to canonical coordinates \(\left ( R,S\right ) \) where the transformation \(\left ( \bar {R},\bar {S}\right ) \rightarrow \left ( R,S+\epsilon \right ) \) can be used.
7. The main goal of Lie symmetry method is to determine \(S,R\). To be able to do this, the quantities \(\xi ,\eta \) must be determined first.
8. The remarkable thing about this method, is that regardless of how complicated the original ode \(\frac {dy}{dx}=\omega \left ( x,y\right ) \) is, if the similarity condition PDE can be solved for \(\xi ,\eta \), then \(R,S\) are found and the ode becomes quadrature \(\frac {dS}{dR}=F\left ( R\right ) \). The ode is then solved in canonical coordinates and the solution transformed back to \(x,y\).
9. The quantity \(\epsilon \) is called the Lie parameter. This is a real quantity which as it goes to zero, gives the identity transformation. In other words, when \(\epsilon =0\) then \(\left ( x,y\right ) =\left ( \bar {x},\bar {y}\right ) \).
10. But there is no free lunch, even in Mathematics. The problem comes down to finding \(\xi ,\eta \). This requires solving a PDE. This is done using ansatz and trial and error. This reason possibly explains why the Lie symmetry method have not become standard in textbooks for solving ODE’s as the algebra and computation needed to find \(\xi ,\eta \) from the PDE becomes very complex to do by hand.
11. Total derivative operator: Given \(f\left ( x,y\right ) \) then \(\frac {df}{dx}=\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}\frac {dy}{dx}\) where it is assumed that \(y\left ( x\right ) \) depends on \(x\). Total derivative operator will be used extensively in all the derivatiations below, so good to practice this. It is written as \(D_{x}=\partial _{x}+\partial _{y}y^{\prime }\) for first order ode, and as \(D_{x}=\partial _{x}+\partial _{y}y^{\prime }+\partial _{y^{\prime }}y^{\prime \prime }\) for second order ode and as \(D_{x}=\partial _{x}+\partial _{y}y^{\prime }+\partial _{y^{\prime }}y^{\prime \prime }+\partial _{y^{\prime \prime }}y^{\prime \prime \prime }\) for third order ode and so on.
12. The notation \(f_{x}\) means partial derivative. Hence \(\frac {\partial f}{\partial x}\) is written as \(f_{x}\). Total derivative will always be written as \(\frac {df}{dx}\). It is important to distinguish between these two as the algebra will get messy with Lie symmetry. Sometimes we write \(f^{\prime }\) to mean \(\frac {df}{dx}\) but it is better to avoid \(f^{\prime }\) and just write \(\frac {df}{dx}\) when \(f\) is function of more than one variable.
13. Given first ode \(\frac {dy}{dx}=\omega \left ( x,y\right ) \), where \(\bar {y}\equiv \bar {y}\left ( x,y\right ) \) and \(\bar {x}\equiv \bar {x}\left ( x,y\right ) \) then then\(\frac {d\bar {y}}{d\bar {x}}\) is given by the following (using the total derivative operator)\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\frac {D_{x}\bar {y}}{D_{x}\bar {x}}\\ & =\frac {\bar {y}_{x}+\bar {y}_{y}y^{\prime }}{\bar {x}_{x}+\bar {x}_{y}y^{\prime }}\\ & =\frac {\bar {y}_{x}+\bar {y}_{y}\omega }{\bar {x}_{x}+\bar {x}_{y}\omega } \end {align*}

14. Given second order ode \(\frac {d^{2}y}{dx^{2}}=\omega \left ( x,y,y^{\prime }\right ) \) where \(\bar {y}\equiv \bar {y}\left ( x,y,y^{\prime }\right ) \) and \(\bar {x}\equiv \bar {x}\left ( x,y,y^{\prime }\right ) \) then \(\frac {d^{2}\bar {y}}{d\bar {x}^{2}}\) is given by\begin {align*} \frac {d^{2}\bar {y}}{d\bar {x}^{2}} & =\frac {D_{x}\frac {d\bar {y}}{d\bar {x}}}{D_{x}\bar {x}}\\ & =\frac {\bar {y}_{x}^{\prime }+\bar {y}_{y}^{\prime }y^{\prime }+\bar {y}_{y^{\prime }}^{\prime }y^{\prime \prime }}{\bar {x}_{x}^{\prime }+\bar {x}_{y}^{\prime }y^{\prime }} \end {align*}

    To simplify notation we used \(\bar {y}^{\prime }\) for \(\frac {d\bar {y}}{d\bar {x}}\) above. The above simplifies to\[ \frac {d^{2}\bar {y}}{d\bar {x}^{2}}=\frac {\bar {y}_{x}^{\prime }+\bar {y}_{y}^{\prime }y^{\prime }+\bar {y}_{y^{\prime }}^{\prime }\omega }{\bar {x}_{x}^{\prime }+\bar {x}_{y}^{\prime }y^{\prime }}\] Keeping in mind that \(\left ( \circ \right ) _{x}\) or \(\left ( \circ \right ) _{y}\) mean partial derivative.

15. Given third order ode \(\frac {d^{3}y}{dx^{3}}=\omega \left ( x,y,y^{\prime },y^{\prime \prime }\right ) \) where \(\bar {y}\equiv \bar {y}\left ( x,y,y^{\prime },y^{\prime \prime }\right ) \) and \(\bar {x}\equiv \bar {x}\left ( x,y,y^{\prime },y^{\prime }\right ) \) then \(\frac {d^{3}\bar {y}}{d\bar {x}^{3}}\) is given by \begin {align*} \frac {d^{3}\bar {y}}{d\bar {x}^{3}} & =\frac {D_{x}\frac {d^{2}\bar {y}}{d\bar {x}^{2}}}{D_{x}\bar {x}}\\ & =\frac {\bar {y}_{x}^{^{\prime \prime }}+\bar {y}_{y}^{\prime \prime }y^{\prime }+\bar {y}_{y^{\prime }}^{\prime \prime }y^{\prime \prime }+\bar {y}_{y^{\prime \prime }}^{\prime \prime }y^{\prime \prime \prime }}{\bar {x}_{x}^{\prime }+\bar {x}_{y}^{\prime }y^{\prime }}\\ & =\frac {\bar {y}_{x}^{^{\prime \prime }}+\bar {y}_{y}^{\prime \prime }y^{\prime }+\bar {y}_{y^{\prime }}^{\prime \prime }y^{\prime \prime }+\bar {y}_{y^{\prime \prime }}^{\prime \prime }\omega }{\bar {x}_{x}^{\prime }+\bar {x}_{y}^{\prime }y^{\prime }} \end {align*}

    To simplify notation we used \(\bar {y}^{\prime \prime }\) for \(\frac {d^{2}\bar {y}}{d\bar {x}^{2}}\) above. And so on for higher order ode’s.