3.1.8 Definitions and various notes

1. 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.
2. \(\omega \left ( x,y\right ) \) is invariant iff \(\boldsymbol {\Gamma }\omega =\xi \left ( x,y\right ) \frac {\partial }{\partial x}+\eta \left ( x,y\right ) \frac {\partial }{\partial y}=0\).
3. 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.
4. 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.
5. 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.
6. 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.
7. 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.
8. 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.
9. 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.
10. A solution curve of \(y^{\prime }=\omega \left ( x,y\right ) \) that maps to itself under the symmetry transformation is called an invariant curve.
11. 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?
12. 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.
13. \(\xi ,\eta \) are called Lie infinitesimal and \(\bar {x},\bar {y}\) are called the Lie group.
14. If we given the \(\xi ,\eta \) then we can find Lie group \(\left ( \bar {x},\bar {y}\right ) \). See example below.
15. 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 \(\eta \left ( x,y\right ) =\left . \frac {\partial }{\partial \epsilon }\bar {y}\right \vert _{\epsilon =0}\).
16. 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.
17. \(\xi ,\eta \) 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 \(\mu \left ( x,y\right ) \).
18. For first order ode, to find Lie infinitesimal, we have to solve first order PDE in 2 variables. For second order ode, to find Lie infinitesimal, we have to solve second order PDE in 3 variables. For third order ode, to find Lie infinitesimal, 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 infinitesimal. 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 familiar 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.