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1.4 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.

  1. Given an ode \(y^{\prime }=\omega \left ( x,y\right ) \) to solve in natural coordinates.

  2. The tangent vector \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) are found. There are two options to find these.

    1. 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

      \[ \bar {x}_{x}\bar {y}_{y}-\bar {x}_{y}\bar {y}_{x}\neq 0 \]

    2. In practice Lie group coordinates \(\left ( \bar {x},\bar {y}\right ) \) are not given and are not known. In this case \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) must be found by solving the similarity condition which results in a PDE (derivation is given below). The PDE for first order ode \(y^{\prime }=\omega \left ( x,y\right ) \) comes out to be

      \[ \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \]
  3. \(\xi ,\eta \) are now used to determine the canonical coordinates \(\left ( X,Y\right ) \). In the canonical coordinates, only \(S\) translation is needed to make the ode quadrature. The transformation is \(\left ( \bar {X},\bar {Y}\right ) \rightarrow \left ( X,Y+\varepsilon \right ) \). This transforms the original ode \(y^{\prime }=\omega \left ( x,y\right ) \) to \(\frac {dY}{dX}=f\left ( X\right ) \) which is then solved by only integration. This is the main advantage of moving to canonical coordinates \(\left ( X,Y\right ) \). In canonical coordinates \(Y\) is like the \(y\) and \(X\) is like the \(x\). i.e. dependent variable is \(Y\left ( X\right ) \) like \(y\left ( x\right ) \) in natural coordinates. The ODE in canonical coordinates is much simpler to solve than in natural coordinates.
  4. The ODE is solved in \(\left ( X,Y\right ) \) space where \(X\equiv X\left ( x,y\right ) ,Y\equiv Y\left ( x,y\right ) \). The transformation from \(\left ( x,y\right ) \) to \(\left ( X,Y\right ) \) is found by solving two set of PDEs using the characteristic method. After finding \(X\left ( x,y\right ) ,Y\left ( x,y\right ) \) the ode will then be given by \(\frac {dY}{dX}=\frac {Y_{x}+Y_{y}\frac {dy}{dx}}{X_{x}+X_{y}\frac {dy}{dx}}\) which will be quadrature. If this ode does not come out as \(\frac {dY}{dX}=f\left ( X\right ) \) then something went wrong in the process. This ode is now solved for \(Y\left ( X\right ) .\) It is the symmetry of the form \(\left ( \bar {X},\bar {Y}\right ) \rightarrow \left ( X,Y+\varepsilon \right ) \) which is of the most interest in the Lie method. This is called a translation transformation along the \(Y\) axis (i.e. vertical). This is because this vertical transformation is what leads to an ode which is solved by just integration.
  5. Solution is transformed from \(Y\left ( X\right ) \) to \(y\left ( x\right ) \).
  6. An alternative to steps (3) to (5) (Which seems to be only applicable to first order odes) 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 hope to add this soon as a backup algorithm if the main one fails. This method is similar to solving exact ode if the integrating factor can be found.
  7. An important property, at least for first order ode’s (I do not know now if this applies to higher order) is that given \(\xi =f\left ( x,y\right ) ,\eta =g\left ( x,y\right ) \), then it is always possible to shift and use \(\xi \equiv 0,\eta =g-\omega f\)  where \(y^{\prime }=\omega \left ( x,y\right ) \). This means everything can be based on \(\xi \equiv 0\) after this shift is done. This can simplify some parts of the computation. Of course 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.

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