Given any first order ODE \begin {equation} \frac {dy}{dx}=\omega \left ( x,y\right ) \tag {A} \end {equation} The first goal is to find a one parameter invariant Lie group transformation that keeps the ode invariant. The Lie parameter the transformation depends on is called \(\epsilon \). This means finding transformation of \(\left ( x,y\right ) \) to new coordinates \(\left ( \bar {x},\bar {y}\right ) \) that keeps the ode the same form when written using \(\bar {x},\bar {y}\).
This view looks at the transformation on the ode itself. Another view is to look at the family of the solution curves of the ode instead. Looking at solution curves transformation is geometrical in nature and can lead to more insight.
What does the transformation mean when looking at solution curves instead of the ODE itself? It is the mapping of a point \(\left ( x,y\right ) \) on one solution curve to another point \(\left ( \bar {x},\bar {y}\right ) \) on another solution curve. If the mapping sends point \(\left ( x,y\right ) \) to another point \(\left ( \bar {x},\bar {y}\right ) \) on the same solution curve, then it is called a trivial mapping or trivial transformation.
As an example, given the ode \(y^{\prime }=0\), this has solutions \(y=c_{1}\). For any constant \(c_{1}\) there is a solution curve. There are infinite number of solution curves. All solution curves are horizontal lines. The mapping \(\left ( x,y\right ) \rightarrow \left ( x+\epsilon ,y\right ) \) is trivial transformation as it moves the point \(\left ( x,y\right ) \) to another point \(\left ( \bar {x},\bar {y}\right ) \) on the same solution curve.
The transformation \(\left ( x,y\right ) \rightarrow \left ( x,y+\epsilon \right ) \) however is non trivial as it moves the point \(\left ( x,y\right ) \) to point \(\left ( \bar {x},\bar {y}\right ) \) on another solution curve. Here \(\bar {x}=x\) and \(\bar {y}=y+\epsilon \). This can also be written \(\left ( x,y\right ) \rightarrow \left ( x,e^{\epsilon }y\right ) \) which is the preferred way.
The transformation \(\left ( x,y\right ) \rightarrow \left ( x+\epsilon ,y+\epsilon \right ) \) is non trivial for this ode. The simplest non trivial transformation that map all points on one solution curve to another solution curve is selected. In canonical coordinates the transformation used has the form \(\left ( R,S\right ) \rightarrow \left ( R,S+\epsilon \right ) \).
Another example is \(y^{\prime }=y\). This has solution curves given by \(y=ce^{x}\). This is a plot showing two such curves for different \(c\) values.
The above shows that a non trivial transformation is given by \(\bar {x}=x+\epsilon ,\bar {y}=y\). This can be found analytically by solving the symmetry condition as will be illustrated below using examples. For this case, the tangent vectors are \(\xi =\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}=1\) and \(\eta =\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0}=0\). In Maple this is found using
ode:=diff(y(x),x)=y(x); DEtools:-symgen(ode) [_xi = 1, _eta = 0]
But the following transformation \(\bar {x}=x,\bar {y}=y+\epsilon \) does not work
This is because it does not leave the ode invariant because \(\frac {d\bar {y}}{d\bar {x}}=\bar {y}\) becomes \(\frac {\bar {y}_{x}+\bar {y}_{y}y^{\prime }}{\bar {x}_{x}+\bar {x}_{y}y^{\prime }}=\bar {y}\), where now \(\bar {y}_{x}=0,\bar {y}_{y}=1,\bar {x}_{x}=1,\bar {x}_{y}=0,\bar {y}=y+\epsilon \), and hence \(\frac {\bar {y}_{x}+\bar {y}_{y}y^{\prime }}{\bar {x}_{x}+\bar {x}_{y}y^{\prime }}=\bar {y}\) simplifies to \(y^{\prime }=y+\epsilon \) which is not the same ode. This shows that \(\bar {x}=x,\bar {y}=y+\epsilon \) is not valid Lie point symmetry.
However \(\bar {x}=x+\epsilon ,\bar {y}=y\) leaves the ODE invariant. In this case \(\bar {y}_{x}=0,\bar {y}_{y}=1,\bar {x}_{x}=1,\bar {x}_{y}=0,\bar {y}=y\) and hence \(\frac {\bar {y}_{x}+\bar {y}_{y}y^{\prime }}{\bar {x}_{x}+\bar {x}_{y}y^{\prime }}=\bar {y}\) becomes \(y^{\prime }=y\) which is the same ode.
The transformation must keep the ode invariant as this is the main definition of symmetry transformation.
In the above, the path the point \(\left ( x,y\right ) \) travels over as it moves to \(\left ( \bar {x},\bar {y}\right ) \) as \(\epsilon \) changes is called the orbit. Each point \(\left ( x,y\right ) \) travels on its orbit during transformation.
In all such transformations, there is a parameter \(\epsilon \) that the transformation depends on. This is why this is called the Lie one parameter symmetry transformation group. There are infinite number of such transformations.
Lie symmetry is called point symmetry, because of the above. It transforms points from an ODE solution curves to points on another solution curves for the same ODE. The identity transformation is when \(\epsilon =0\), since then the point is transformed to itself.
An example using an ODE. The Clairaut ode of the form \(y=xf\left ( p\right ) +g\left ( p\right ) \) where \(p\equiv y^{\prime }\). \begin {align} x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }+m & =0\tag {1}\\ y & =x\frac {\left ( y^{\prime }\right ) ^{2}}{m}+y\frac {y^{\prime }}{m}\nonumber \end {align}
Where \(f\left ( p\right ) =\frac {\left ( y^{\prime }\right ) ^{2}}{m}\) and \(g\left ( p\right ) =\frac {y^{\prime }}{m}\). Using the dilation transformation Lie group\begin {align} \bar {x} & \equiv \bar {x}\left ( x,y;\epsilon \right ) =e^{2\epsilon }x\tag {2}\\ \bar {y} & \equiv \bar {y}\left ( x,y;\epsilon \right ) =e^{\epsilon }y \tag {3} \end {align}
Eq. (1) is now expressed in the new coordinates \(\bar {x},\bar {y}\,\). If this results in same same ode form but written in \(\bar {x},\bar {y}\,\) then the transformation is invariant. But how to find \(\frac {d\bar {y}}{d\bar {x}}\) ? This is done as follows\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\frac {\frac {d\bar {y}}{dx}}{\frac {d\bar {x}}{dx}}\\ & =\frac {\bar {y}_{x}+\bar {y}_{y}\frac {dy}{dx}}{\bar {x}_{x}+\bar {x}_{y}\frac {dy}{dx}} \end {align*}
In this example \(\bar {y}_{x}=0,\bar {y}_{y}=e^{\epsilon },\bar {x}_{x}=e^{2\epsilon },\bar {x}_{y}=0.\) The above now becomes\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\frac {e^{\epsilon }\frac {dy}{dx}}{e^{2\epsilon }}\\ & =e^{-\epsilon }\frac {dy}{dx} \end {align*}
Writing (1) in terms of \(\bar {x},\bar {y}\) now gives\begin {align} \bar {x}\left ( \frac {d\bar {y}}{d\bar {x}}\right ) ^{2}-\bar {y}\frac {d\bar {y}}{d\bar {x}}+m & =0\tag {4}\\ \left ( e^{2\epsilon }x\right ) \left ( e^{-\epsilon }\frac {dy}{dx}\right ) ^{2}-\left ( e^{\epsilon }y\right ) e^{-\epsilon }\frac {dy}{dx}+m & =0\nonumber \\ x\left ( \frac {dy}{dx}\right ) ^{2}-y\frac {dy}{dx}+m & =0 \tag {5} \end {align}
Which gives the same ode. The above method starts by replacing the given ode by \(\bar {x},\bar {y},\frac {d\bar {y}}{d\bar {x}}\) and finds if the result gives back the original ode in \(x,y,\frac {dy}{dx}\). This is simpler than having to transform the original ode to \(\bar {x},\bar {y},\frac {d\bar {y}}{d\bar {x}}\). This transformation can be verified in Maple as follows
ode:=x*diff(y(x),x)^2-y(x)*diff(y(x),x)+m=0; the_tr:={x=X*exp(-2*s),y(x)=Y(X)*exp(-s)}; newode:=PDEtools:-dchange(the_tr,ode,{Y(X),X},'known'={y(x)},'uknown'={Y(X)}); diff(Y(X), X)^2*X - Y(X)*diff(Y(X), X) + m = 0
Comparing (4) to (5) shows that the ode form did not change, only the letters changed from \(x\) to \(\bar {x}\) and \(y\) to \(\bar {y}\). The resulting ode must never have the parameter \(\epsilon \) show or remain in it.
The above shows how to verify that a transformation is invariant or not. In Lie group transformation there is only one parameter \(\epsilon \) and the transformation is obtained by evaluating the group as \(\epsilon \) goes to zero.
But how does this help in solving the ode? If the ode in \(x,y\) is hard to solve, then the ode written with \(\bar {x},\bar {y}\,\) will also be hard to solve since it is the same. But Eq. (4) is not what is used to solve the ode, but the above is just to verify the transformation is invariant. Similarity transformation is used to determine tangent vectors \(\xi ,\eta \) only. Then the ode in canonical coordinates is used instead. In the canonical coordinates \(\left ( R,S\right ) \) the ode becomes quadrature and solved by integration. The transformation found above is only one step toward finding \(\left ( R,S\right ) \) and it is these canonical coordinates that are the goal and not \(\bar {x},\bar {y}\).