2.2.22.3 Outline of the steps in solving a differential equation using Lie symmetry
method
These are the steps in solving an ODE using Lie symmetry method.
Given an ode \(y^{\prime }=\omega \left ( x,y\right ) \) to solve in natural coordinates.
Now the tangent vector \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) are found. There are two options.
If Lie group coordinates \(\left ( \bar {x},\bar {y}\right ) \) are given, then it is easy to determine \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) using\begin {align*} \xi \left ( x,y\right ) & =\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}\\ \eta \left ( x,y\right ) & =\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0} \end {align*}
Lie group coordinates \(\left ( \bar {x},\bar {y}\right ) \) must also satisfy \[ \bar {x}_{x}\bar {y}_{y}-\bar {x}_{y}\bar {y}_{x}\neq 0 \]
In practice Lie group coordinates \(\left ( \bar {x},\bar {y}\right ) \) are not given and are not known. In this case \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \)
are found by solving the similarity condition which results in a PDE (derivation is
given below). The PDE is \[ \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \]
\(\xi ,\eta \) are now used to determine the canonical coordinates \(\left ( R,S\right ) \). In the canonical coordinates,
only \(S\) translation is needed to make the ode quadrature. The transformation is \(\left ( R,S\right ) \rightarrow \left ( R,S+\varepsilon \right ) \). This
transforms the original ode \(y^{\prime }=\omega \left ( x,y\right ) \) to \(\frac {dS}{dR}=F\left ( R\right ) \) which is then solved by only integration. This is the
main advantage of moving to canonical coordinates \(\left ( R,S\right ) \).
The ODE is solved in \(\left ( R,S\right ) \) space where \(R\equiv R\left ( x,y\right ) ,S\equiv S\left ( x,y\right ) \). The transformation from \(\left ( x,y\right ) \) to \(\left ( R,S\right ) \) is found by solving
two set of PDEs using the characteristic method. After finding \(R\left ( x,y\right ) ,S\left ( x,y\right ) \) the ode will then
be given by \(\frac {dS}{dR}=\frac {S_{x}+S_{y}\frac {dy}{dx}}{R_{x}+R_{y}\frac {dy}{dx}}\) which will be quadrature. If this ode does not come out as \(\frac {dS}{dR}=F\left ( R\right ) \)
then something went wrong in the process. This ode is now solved for \(S\left ( R\right ) .\) It is
the symmetry of the form \(\left ( R,S\right ) \rightarrow \left ( R,S+\varepsilon \right ) \) which is of the most interest in the Lie method.
This is called a translation transformation along the \(y\) axis (or the \(S\) axis).
This is because this transformation leads to an ode which is solved by just
integration.
Transform the solution from \(S\left ( R\right ) \) to \(y\left ( x\right ) \).
An alternative to steps (3) to (5) is to use \(\xi ,\eta \) to determine an integrating factor \(\mu \left ( x,y\right ) \) which is
given by \(\mu \left ( x,y\right ) =\frac {1}{\eta -\xi \omega }\) then the general solution to \(y^{\prime }=\omega \left ( x,y\right ) \) can be written directly as \(\int \mu \left ( x,y\right ) \left ( dy-\omega dx\right ) =c_{1}\) or \(\int \frac {dy-\omega dx}{\eta -\xi \omega }=c_{1}\) but this requires
finding a function \(F\left ( x,y\right ) \) whose differential is \(dF=\frac {dy-\omega dx}{\eta -\xi \omega }\) and now the solution becomes \(\int dF=c_{1}\,\) or \(F=c_{1}\). If we can
integrate this using \(\int \mu dy-\int \mu \omega dx=c_{1}\) then this is the solution to the ode. It is implicit in \(y\left ( x\right ) \). Currently my
program does not implement Lie symmetry to find an integrating factor due to
difficulty of finding \(dF\) that satisfies\(\ dF=\frac {dy-\omega dx}{\eta -\xi \omega }\) or in carrying out the integration in all general
cases but I hope to add this soon as a backup algorithm if the main one
fails.
An important property, at least for first order ode’s (I do not know now if this carries
to higher order) is that given \(\xi =f\left ( x,y\right ) ,\eta =g\left ( x,y\right ) \), then we can always shift and use \(\xi \equiv 0,\eta =g-\omega f\) where \(y^{\prime }=\omega \left ( x,y\right ) \). This means
we can always base everything on \(\xi \equiv 0\) after this shift is done to \(\eta \). This can simplify some
parts of the computation. Ofcourse if \(\xi \) was found to be zero initially, i.e. just
after solving the linearized similarity PDE, then there is nothing more to
do.
The most difficult step in all of the above is 2(b) which requires finding \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \). In practice Lie
group \(\bar {x},\bar {y}\) transformation is not given. Lie infinitesimal \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) have to be found directly from the
linearized symmetry condition PDE using ansatz and by trial and error. The following
diagram illustrates the above steps.
The following diagram illustrates the above steps when we carry the shifting step in order to
force \(\xi =0\). We see that It simplifies the algorithm as now we can just assume \(\xi =0\) and we do not
have to check for different cases as before.