3.4.6 Moving to canonical coordinates \(R,S\)

The next step is to determine what is called the canonical coordinates \(\left ( R,S\right ) \). In these canonical coordinates the ODE becomes a quadrature and solved by integration. Once solved, the solution is transformed back to \(\left ( x,y\right ) \). The canonical coordinates \(\left ( R,S\right ) \) are found as follows. Selecting the transformation to be

\begin{align} \bar {R} & =R\tag {15}\\ \bar {S} & =S+\epsilon \tag {16}\end{align}

Eq. (15) becomes

\[ \left . \frac {\partial \bar {R}}{\partial \epsilon }\right \vert _{\epsilon =0}=\left . \left ( \frac {\partial \bar {R}}{\partial x}\frac {dx}{d\epsilon }\right ) \right \vert _{\epsilon =0}+\left . \left ( \frac {\partial \bar {R}}{\partial y}\frac {dy}{d\epsilon }\right ) \right \vert _{\epsilon =0}\]

But \(\left . \frac {\partial \bar {R}}{\partial x}\right \vert _{\epsilon =0}=\frac {\partial R}{\partial x}\) and \(\left . \frac {dx}{d\epsilon }\right \vert _{\epsilon =0}=\xi \left ( x,y\right ) \) and similarly \(\left . \frac {\partial \bar {R}}{\partial y}\right \vert _{\epsilon =0}=\frac {\partial R}{\partial y}\) and \(\left . \frac {dy}{d\epsilon }\right \vert _{\epsilon =0}=\eta \left ( x,y\right ) \). The above becomes

\[ \left . \frac {\partial \bar {R}}{\partial \epsilon }\right \vert _{\epsilon =0}=\frac {\partial R}{\partial x}\xi +\frac {\partial R}{\partial y}\eta \]

But \(\left . \frac {\partial \bar {R}}{\partial \epsilon }\right \vert _{\epsilon =0}=0\) since \(\bar {R}=R\). The above reduces to

\[ 0=\frac {\partial R}{\partial x}\xi +\frac {\partial R}{\partial y}\eta \]

This PDE have solution using symmetry method given by

\begin{align} \frac {dR}{dt} & =0\tag {15A}\\ \frac {dx}{dt} & =\xi \tag {15B}\\ \frac {dy}{dt} & =\eta \tag {15C}\end{align}

The same procedure is applied to Eq. (16) which gives

\[ \left . \frac {\partial \bar {S}}{\partial \epsilon }\right \vert _{\epsilon =0}=\left . \left ( \frac {\partial \bar {S}}{\partial x}\frac {dx}{d\epsilon }\right ) \right \vert _{\epsilon =0}+\left . \left ( \frac {\partial \bar {S}}{\partial y}\frac {dy}{d\epsilon }\right ) \right \vert _{\epsilon =0}\]

But \(\left . \frac {\partial \bar {S}}{\partial x}\right \vert _{\epsilon =0}=\frac {\partial S}{\partial x}\) and \(\left . \frac {dx}{d\epsilon }\right \vert _{\epsilon =0}=\xi \left ( x,y\right ) \) and similarly \(\left . \frac {\partial \bar {S}}{\partial y}\right \vert _{\epsilon =0}=\frac {\partial S}{\partial y}\) and \(\left . \frac {dy}{d\epsilon }\right \vert _{\epsilon =0}=\eta \left ( x,y\right ) \,\). The above becomes

\[ \left . \frac {\partial \bar {S}}{\partial \epsilon }\right \vert _{\epsilon =0}=\frac {\partial R}{\partial x}\xi +\frac {\partial R}{\partial y}\eta \]

But \(\left . \frac {\partial \bar {S}}{\partial \epsilon }\right \vert _{\epsilon =0}=1\) since \(\bar {S}=S+\epsilon \). The above reduces to

\[ 1=\frac {\partial S}{\partial x}\xi +\frac {\partial S}{\partial y}\eta \]

This PDE have solution using symmetry method given by

\begin{align} \frac {dS}{dt} & =1\tag {16A}\\ \frac {dx}{dt} & =\xi \tag {16B}\\ \frac {dy}{dt} & =\eta \tag {16C}\end{align}

Equations (15A,B,C) are used to solve for \(R\left ( x,y\right ) \) and equations (16A,B,C) are used to solve for \(S\left ( x,y\right ) \). Starting with \(R\). In the case when \(\xi =0\) the equations become

\begin{align*} \frac {dR}{dt} & =0\\ \frac {dx}{dt} & =0\\ \frac {dy}{dt} & =\eta \end{align*}

First equation above gives \(R=c_{1}\). Second equation gives \(x=c_{2}\).  Letting \(c_{1}=c_{2}\) then

\[ R=x \]

If \(\xi \neq 0\) then combining Eqs. (15B,15C) gives  

\begin{align*} \frac {dy}{dx} & =\frac {\eta }{\xi }\\ R & =c_{1}\end{align*}

The ODE \(\frac {dy}{dx}=\frac {\eta }{\xi }\) is solved first and the constant of integration is replaced by \(R\). Hence \(R\) is now found. \(S\left ( x,y\right ) \) is found similarly using Eqs. (16A,B,C).  If \(\xi =0\) then

\begin{align*} \frac {dS}{dt} & =1\\ \frac {dx}{dt} & =0\\ \frac {dy}{dt} & =\eta \end{align*}

The first and third equations give

\begin{align*} \frac {dS}{dy} & =\frac {1}{\eta }\\ S & =\int \frac {1}{\eta }dy \end{align*}

If \(\xi \neq 0\) then using the second and third equation gives

\begin{align*} \frac {dS}{dx} & =\frac {1}{\xi }\\ S & =\int \frac {1}{\xi }dx \end{align*}

Now that \(R,S\) are found and the problem is solved. The ode in \(\left ( R,S\right ) \) space is set up using

\begin{equation} \frac {dS}{dR}=\frac {S_{x}+S_{y}\frac {dy}{dx}}{R_{x}+R_{y}\frac {dy}{dx}} \tag {16}\end{equation}

Where \(\frac {dy}{dx}=\omega \left ( x,y\right ) \) which is given. The solution \(S\left ( R\right ) \) is next converted back to \(y\left ( x\right ) \).

Examples below illustrate how this done on a number of ODE’s. Eq. (16) is solved by quadrature. This is the whole point of Lie symmetry method, is that the original ode is solved in canonical coordinates where it is much easier to solve and the solution is transformed back to natural coordinates.

The only way to understand this method well, is to workout some problems. To learn more about the theory of Lie transformation itself and why it works, there are many links in my links page on the subject.