3.1.2 Terminology used and high level introduction

1. Sophus Lie was inspired by Galois theory for solving algebraic equations.

2. The goal of Lie symmetry for solving first order ode is to transform the ode \(\frac {dy}{dx}=\omega \left ( x,y\right ) \) to canonical coordinates \(\frac {dY}{dX}=F\left ( X\right ) \) where it is easily solved by quadrature. Once the solution \(Y\left ( X\right ) \) is found, it is converted back to \(y\left ( x\right ) \) in the original \(x,y\) coordinates, thus obtaining the solution \(y\left ( x\right ) \) to the original ode. This method works regardless of how complicated the original ode happened to be (linear or not). But this requires finding what is called the Lie infinitesimals \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) which requires solving a PDE using ansatz and this can be difficult. Different algorithms are designed to help find \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) using different forms of ansatz.
   

   

   Figure 3.2: Lie symmetry for solving first order ode

3. \(x,y\) are the natural coordinates used in the input ode \(\frac {dy}{dx}=\omega \left ( x,y\right ) \). For example \(y^{\prime }=x^{2}+y^{2}\). Here \(\omega =x^{2}+y^{2}\).
4. \(\bar {x},\bar {y}\) are called the Lie group (local) transformation coordinates. The Lie transformation is one parameter transformation. Meaning it depends only on one parameter. Some books call this \(\lambda \) and some call it \(\varepsilon \). Here \(\varepsilon \) is used. Hence we write \(\left ( \bar {x},\bar {y}\right ) =T\left ( x,y;\varepsilon \right ) \) to mean transformation \(T\) is applied on point \(\left ( x,y\right ) \) to obtain new point \(\left ( \bar {x},\bar {y}\right ) \) and this transformation depends on the value used for \(\varepsilon \). The parameter \(\varepsilon \) is real value. The ode \(\frac {d\bar {y}}{d\bar {x}}=\omega \left ( \bar {x},\bar {y}\right ) \) must remain invariant (i.e. same shape) but with new letters \(\bar {x},\bar {y}\) that replace each of the letters \(x,y\) in the original ode. If the new ode does not have same exact form, then this is not valid Lie symmetry transformation that was used.
5. The group transformation is defined as \(T_{\varepsilon }:\left \{ \bar {x}=\varphi \left ( x,y,\varepsilon \right ) ,\bar {y}=\psi \left ( x,y,\varepsilon \right ) \right \} \). It is required that \(\varphi \left ( x,y,\varepsilon \right ) ,\psi \left ( x,y,\varepsilon \right ) \) are independent of each others, which means the Jacobian do not vanish. Hence \(\begin {vmatrix} \varphi _{x} & \varphi _{y}\\ \psi _{x} & \psi _{y}\end {vmatrix} \neq 0\). Lie group comes from the above transformation group when expanding \(\varphi \left ( x,y,\varepsilon \right ) ,\psi \left ( x,y,\varepsilon \right ) \) in Taylor series near \(\left ( x,y\right ) \) and keep linear terms, which results in \(\bar {x}\approx x+\varepsilon \xi \left ( x,y\right ) ,\bar {y}\approx y+\varepsilon \eta \left ( x,y\right ) \).

6. If given Transformation group \(T_{\varepsilon }\) defined as \(\bar {x}=\varphi \left ( x,y,\varepsilon \right ) ,\bar {y}=\psi \left ( x,y,\varepsilon \right ) \), then \(\xi ,\eta \) are found as follows.

   \begin{align*} \xi & =\left . \frac {d\bar {x}}{d\varepsilon }\right \vert _{\varepsilon =0}\\ \eta & =\left . \frac {d\bar {y}}{d\varepsilon }\right \vert _{\varepsilon =0}\end{align*}

   For example. Given transformation \(\bar {x}=x\cos \left ( \varepsilon \right ) +y\sin \left ( \varepsilon \right ) ,\bar {y}=y\cos \left ( \varepsilon \right ) -x\sin \left ( \varepsilon \right ) \), then the above gives \(\xi =y,\eta =-x\) and the Lie transformation becomes

   \begin{align*} \bar {x} & \approx x+\varepsilon y\\ \bar {y} & \approx y-\varepsilon x \end{align*}
7. The quantities \(\left ( \xi ,\eta \right ) \) define the tangent direction at \(\left ( x,y\right ) \) of the path that is taken to move \(\left ( x,y\right ) \) to \(\left ( \bar {x},\bar {y}\right ) \). In other words, starting from any point \(\left ( x,y\right ) \) and calculating \(\left ( \xi ,\eta \right ) \) at \(\left ( x,y\right ) \), then the line going from \(\left ( x,y\right ) \) to the point \(\left ( x+\varepsilon \xi ,y+\varepsilon \eta \right ) \) for a very small \(\varepsilon \) value, will be tangent line to the path from \(\left ( x,y\right ) \) to \(\left ( \bar {x},\bar {y}\right ) \).
8. It is good to look at \(\bar {x}=x+\varepsilon \xi \left ( x,y\right ) \) as in kinematics, where \(x=x\left ( 0\right ) +v_{x}t\), where now \(\varepsilon \) represents the time and \(x\left ( 0\right ) \) is initial position and \(x\) is final position and \(\xi \left ( x,y\right ) \) is the speed \(v_{x}\) which is a function of position. Same for the \(y\) coordinate. \(y=y\left ( 0\right ) +v_{y}t\). In this view, as \(t\) increases the point moves more and all points covered in the path are the orbit of point \(\left ( x,y\right ) \).

9. The coordinates \(\left ( X,Y\right ) \) (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. In other words, in canonical coordinates, Lie transformation is given by

   \begin{align*} \bar {X} & =X\\ \bar {Y} & =Y+\varepsilon \end{align*}

   Comparing with the original coordinates which is

   \begin{align*} \bar {x} & \approx x+\varepsilon \xi \\ \bar {y} & \approx y+\varepsilon \eta \end{align*}

   The ode is always solved in canonical coordinates \(\left ( X,Y\right ) \) and not in \(\left ( x,y\right ) \) since it is much simpler to solve it in those coordinates.

10. \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) are called the Lie infinitesimals, also called tangent vectors. They are functions of \(\left ( x,y\right ) \). These are the core quantities of Lie symmetry method. These can be calculated knowing \(\bar {x},\bar {y}\). Also \(\bar {x},\bar {y}\) can be calculated given \(\xi ,\eta \). In practice, \(\bar {x},\bar {y}\) are not given, and hence these have to be found using solving a PDE. It is \(\xi ,\eta \) which are the most important quantities that need to be determined in order to find the canonical coordinates \(X,Y\).
11. The tangent vectors \(\xi ,\eta \) are calculated at \(\epsilon =0\). They are defined as \(\xi =\left . \frac {dx}{d\epsilon }\right \vert _{\epsilon =0},\eta =\left . \frac {dy}{d\epsilon }\right \vert _{\epsilon =0}\). The point \(\left ( \bar {x},\bar {y}\right ) \) (orbit of \(\left ( x,y\right ) \)) is given by \(\bar {x}=x+\xi \epsilon \) and \(\bar {y}=y+\eta \epsilon \).

12. Given

    \begin{align*} \bar {x} & \equiv \bar {x}\left ( x,y;\epsilon \right ) \\ \bar {y} & \equiv \bar {y}\left ( x,y;\epsilon \right ) \end{align*}

    Expanding using Taylor series near\(\ \epsilon =0\) gives (see sections below for more details)

    \begin{align*} \bar {x} & =x+\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}\epsilon +O\left ( \epsilon ^{2}\right ) \\ & =x+\epsilon \xi \left ( x,y\right ) \\ \bar {y} & =y+\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0}\epsilon +O\left ( \epsilon ^{2}\right ) \\ & =y+\epsilon \eta \left ( x,y\right ) \end{align*}

    The above shows the importance of \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \). These (along with the specific value of \(\epsilon \)) determine the orbit of each point \(\left ( x,y\right ) \).

13. The orbit of a point \(A\) given by natural coordinates \(\left ( x,y\right ) \) is the set of all possible points \(\left ( \bar {x},\bar {y}\right ) \) that the point \(A\) transforms to for all possible value of \(\varepsilon \).
    

    

    Figure 3.3: The orbit of point \(A\) using Lie transformation

14. The ultimate goal is write \(\frac {dy}{dx}=\omega \left ( x,y\right ) \) in \(X,Y\) coordinates where symmetry have the ideal form \(\left ( \bar {X},\bar {Y}\right ) =\left ( X,Y+\varepsilon \right ) \) because this leads to ode of form \(\frac {dY}{dX}=f\left ( X\right ) \). The right hand side should always be a function of \(X\) only in canonical coordinates.
15. The ideal transformation has the form \(\left ( \bar {X},\bar {Y}\right ) \rightarrow \left ( X,Y+\varepsilon \right ) \) as mentioned above, because with this transformation the ode becomes quadrature in the transformed coordinates. But because not all ode’s have this transformation as given, the ode is first transformed to canonical coordinates \(\left ( X,Y\right ) \) where the transformation is \(\left ( \bar {X},\bar {Y}\right ) \rightarrow \left ( X,Y+\varepsilon \right ) \) is imposed. If the transformation \(\left ( \bar {x},\bar {y}\right ) \rightarrow \left ( x,y+\epsilon \right ) \) is already present in original coordinates, then there will be no need for canonical coordinates \(\left ( X,Y\right ) \).
16. The main goal of Lie symmetry method is to determine \(X,Y\) and solve the ode \(\frac {dY}{dX}=f\left ( X\right ) \) in that space instead in the natural coordinates \(\left ( x,y\right ) \). To be able to do this, the quantities \(\xi ,\eta \) must be determined first.
17. The remarkable thing about Lie symmetry 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 \(X,Y\) can always be found and the ode becomes quadrature \(\frac {dY}{dX}=f\left ( X\right ) \). The ode is then solved in canonical coordinates and the solution transformed back to \(x,y\).
18. 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 \(\left ( x,y\right ) =\left ( \bar {x},\bar {y}\right ) \).
19. 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 become very complex to do by hand.
20. 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 derivations 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.
21. 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.

22. 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*}

23. 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 have used \(\bar {y}^{\prime }\) for \(\frac {d\bar {y}}{d\bar {x}}\) in the 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.

24. 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.