infinitesimal generator operator. \(\boldsymbol {\Gamma }=\xi \left ( x,y\right ) \frac {\partial }{\partial x}+\eta \left ( x,y\right ) \frac {\partial }{\partial y}\). Any first order ode has such generator. For
instance, for the ode \(y^{\prime }=\omega \left ( x,y\right ) \) then \(\boldsymbol {\Gamma }\omega =\xi \frac {\partial \omega }{\partial x}+\eta \frac {\partial \omega }{\partial y}\). The ode \(y^{\prime }=\omega \left ( x,y\right ) =\frac {y}{x}+x\) has solution \(y=x^{2}+xc_{1}\), therefore the solution family
is \(\phi \left ( x,y\right ) =\frac {y-x^{2}}{x}=c\). Using \(\xi =0,\eta =x\) then \(\boldsymbol {\Gamma }\phi =x\frac {\partial \left ( \frac {y-x^{2}}{x}\right ) }{\partial y}=1\). This is another example: using \(\xi =x,\eta =2y\,\), hence \(\boldsymbol {\Gamma }\phi =x\frac {\partial \left ( \frac {y-x^{2}}{x}\right ) }{\partial x}+2y\frac {\partial \left ( \frac {y-x^{2}}{x}\right ) }{\partial y}=x\left ( -\frac {y}{x^{2}}-1\right ) +2y\left ( \frac {1}{x}\right ) =-\frac {y}{x}-1+2\frac {y}{x}=\frac {y}{x}-1\neq 1\). I must be not applying
the symmetry generator correct as the result supposed to be \(1\). Need to visit this
again. See book Bluman and Anco, page 109. Maybe some of the assumptions
for using this generator are not satisfied for this ode.
The linearized PDE from the symmetry condition is \(\omega \xi _{x}+\omega ^{2}\xi _{y}+\omega _{x}\xi =\omega _{y}\eta +\eta _{x}+\omega \eta _{y}\). This is used to determine
tangent vector \(\left ( \xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \right ) \) which is one of the core parts of the algorithm to solve the ode
using symmetry methods. There are infinite number of solutions and only one is
needed.
Symmetries and first integrals are the two most important structures of
differential equations. First integral is quantity that depends on \(x,y\) and when
integrated over any solution curve is constant.
Lie symmetry allows one to reduce the order of an ode by one. So if we have third
order ode and we know the symmetry for it, we can change the ode to second
order ode. Then if apply the symmetry for this second order ode, its order is
reduced to one now.
If \(\xi ,\eta \) are known then the canonical coordinates \(R,S\) can now be found as functions of \(x,y\).
We just \(\xi ,\eta \) to find \(R,S\). Once \(R,S\) are known then \(\frac {dS}{dR}=f\left ( R\right ) \) can be formulated. This ode is solved for
\(S\) by quadrature. Final solution is found by replacing \(R,S\) back by \(x,y\). I have functions
and a solver now written and complete to do all of this but just for first order
ode’s only. I need to start on second order ode’s after that. The main and most
difficult step is in finding \(\xi ,\eta \). Currently I only use multivariable polynomial ansatz
up to second order for \(\xi \) and multivariable polynomial ansatz up to third order
for \(\eta \) and then try all possible combinations. This is not very efficient. But works
for now. I need to add better and more efficient methods to finding \(\xi ,\eta \) but need to
do more research on this.
When using polynomial ansatz to find \(\xi ,\eta \) do not mix \(x,y\) in both ansatz. For example
if we use \(\xi =p\left ( x\right ) \) then can use \(\eta =q\left ( x\right ) \) or \(\eta =q\left ( x,y\right ) \) polynomial ansatz to find \(\eta \). But do not try \(\xi =p\left ( x,y\right ) \) ansatz
with \(\eta =q\left ( x,y\right ) \) ansatz. In other words, if one ansatz polynomial is multivariable, then the
other should be single variable. Otherwise results will be complicated and this
defeats the whole ides of using Lie symmetry as the ode generated will be as
complicated or more than the original ode we are trying to solve. I found this
the hard way. I was generating all permutations of \(\xi ,\eta \) ansatz’s but with both as
multivariable polynomials. This did not work well.
Symmetries on the ode itself, is same as talking about symmetries on solution
curves. i.e. given an ode \(y^{\prime }=\omega \left ( x,y\right ) \) with solution \(y=f\left ( x\right ) \), then when we look for symmetry on the
ode which leaves the ode looking the same but using the new variables \(\bar {x},\bar {y}\). This
is the same as when we look for symmetry which maps any point \(\left ( x,y\right ) \) on solution
curve \(y=f\left ( x\right ) \) to another solution curve. In other words, the symmetry will map all
solution curves of \(y^{\prime }=\omega \left ( x,y\right ) \) to the same solution curves. i.e. a specific solution curve \(y=f\left ( x,c_{1}\right ) \) will
be mapped to \(y=f\left ( x,c_{2}\right ) \). All solution curves of \(y^{\prime }=\omega \left ( x,y\right ) \) will be mapped to the same of solution
curves. But each curve maps to another curve within the same set. If the same
curve maps to itself, then this is called invariant curve.
An orbit is the name given to the path the transformation moves the point \(\left ( x,y\right ) \)
from one solution curve to another point on another solution curve due to the
symmetry transformation.
A solution curve of \(y^{\prime }=\omega \left ( x,y\right ) \) that maps to itself under the symmetry transformation is
called an invariant curve.
Not every first order ode has symmetry. At least according to Maple. For example
\(y^{\prime }+y^{3}+xy^{2}=0\) which is Abel ode type, it found no symmetries using way=all. May be with
special hint it can find symmetry?
After trying polynomials ansatz, I find it is limited. Since it will only find
symmetries that has polynomials form. A more powerful ansatz is the functional
form. But these are much harder to work with but they are more general at
same time and can find symmetries that can’t be found with just polynomials. So
I have to learn how to use functional ansatz’s. Currently I only use Polynomials.
\(\xi ,\eta \) are called Lie infinitesimal and \(\bar {x},\bar {y}\) are called the Lie group.
If we given the \(\xi ,\eta \) then we can find Lie group \(\left ( \bar {x},\bar {y}\right ) \). See example below.
If we are given Lie group \(\left ( \bar {x},\bar {y}\right ) \) then we can find the infinitesimal using \(\xi \left ( x,y\right ) =\left . \frac {\partial }{\partial \epsilon }\bar {x}\right \vert _{\epsilon =0}\) and \(\eta \left ( x,y\right ) =\left . \frac {\partial }{\partial \epsilon }\bar {y}\right \vert _{\epsilon =0}\).
First order ode have infinite number of symmetries. Talking about symmetry of
an ode is the same as talking about symmetry between solution curves of the
ode itself. i.e. symmetry then becomes finding mapping that maps each solution
curve to another one in the same family of solutions of the ode.
\(\xi ,\eta \) can also be used to find the integrating factor for the first order ode. This is
given by \(\mu \left ( x,y\right ) =\frac {1}{\eta -\xi \omega }\) where the ode is \(y^{\prime }\left ( x\right ) =\omega \left ( x,y\right ) \,\). This gives an alternative approach to solve the ode.
I still need to add examples using \(\mu \left ( x,y\right ) \).
For first order ode, to find Lie infinitesmilas, we have to solve first order PDE in
2 variables. For second order ode, to find Lie infinitesmilas, we have to solve
second order PDE in 3 variables. For third order ode, to find Lie infinitesmilas,
we have to solve third order PDE in 4 variables and so on. Hence in general, for
\(n^{th}\) order ode, we have to solve \(n^{th}\) order PDE in \(n+1\) variables to find the required Lie
infinitesmilas. For first order, these variables are \(\xi ,\eta \) and the PDE is \(\eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\). Currently my
program only handles first order odes. Once I am more familar with Lie method
for second order ode, will update these notes. See at the end a section on just
second order ode that I started working on.